ANALYSIS  OF  STATICALLY  INDETERMINATE  BUILDINC 
BY  THE  APPLICATION  OF  MAXWELL’S  THEOREM 
OF  RECIPROCAL  DISPLACEMENTS 


JOHN  WILLIAM  ROWLEY 

B.  S.  University  of  Missouri,  1921 


THESIS 

SUBMITTED  IN  PARTIAL  FULFILLMENT  OF  THE  REQUIREMENTS 
FOR  THE  DEGREE  OF  MASTER  OF  SCIENCE  IN  CIVIL 
ENGINEERING  IN  THE  GRADUATE  SCHOOL 
OF  THE  UNIVERSITY  OF  ILLINOIS, 

1922 


URBANA,  ILLINOIS 


r M 


Digitized  by  the  Internet  Archive 
in  2015 


https://archive.org/details/analysisofstaticOOrowl 


/37ai3  tAp 


1^22 

R79 

UNIVERSITY  OF  ILLINOIS 

THE  GRADUATE  SCHOOL 


1922 


i HEREBY  RECOMMEND  THAT  THE  THESIS  PREPARED  UNDER  MY 


SUPERVISION  BY_ 


JQ.H1L. WILLIAM  ROWLEY 


ENTITLEDALALY-S1S_0F_ 


: _BU  ILDIIIG  5MTS 


BY_THE  APPLICATIQH  QF-MAXY/ELLJ-a-  THEORF.I 
PLACEMENTS 

BE  ACCEPTED  AS  FULFILLING  THIS  PART  OF  THE  REQUIREMENTS  FOR 


THE  DEGREE 


_QF_  SCIENCE  I2I  CIYIL  ENGINEERIIIG 


In  Charge  of  Thesis 


Head  of  Department 


Recommendation  concurred  in* 


Committee 


on 


Final  Examination* 


•Required  for  doctor’s  degree  but  not  for  master's 


509405 


I 


TABLE  OF  COIJTEI^TS 


I.  INTROBUCTIOU 

Pag 

!•  Preliminary. 1 

2*  Scope  and  Purpose 4 

3.  Acknowledgements • • 4 

II.  OUTLm  OP  METHOD 


4.  General  Application  of  Maxwell’s  Theorem 


of  Reciprocal  Displacements  6 

5.  Application  to  Class  I, Columns  Hinged  at 

Their  Bases 8 

Case  A,  Horizontal  Reactions 8 

6.  Application  to  Class  II, Columns  Fixed  at 

Their  Bases.  15 

Case  A, Horizontal  Reactions  • ......  15 

Case  B,  Vertical  Reactions 20 

Case  C,  Resisting  Moments  at  the  Bases  of 
the  Columns 26 


III.  APPLICATION  TO  THE  DETSRIIIMTION  OF  REACTIONS 

7.  Determination  of  Reactions  Due  to  Vertical 

Loads 33 

8.  Determination  of  Reactions  Due  to  Comhined 

Wind  and  Vertical  Loads.  36 

9.  Bents  Analyzed 39 

10.  Results. 40 

IV.  TEST  OF  PAPER  MODEL  OF  BENT 

11.  Description  of  Test .42 

12.  Results  of  Test  45 


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V,  DISCUSSIOIT  MD  CONCLUSIONS 

Page 

13*  Effect  of  Variations  in  Different  Characteristics 

of  the  Bent  on  Reactions  Due  to  Vertical  Loads.  .47 

14.  Errors  in  Connnon  Assumptions  for  Various 

Combinations  of  Wind  and  Vertical  Loads.  ...  .50 

15.  Conclusions 53 


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LIST  OF  TABLES 

VERTICAL  DEPLECTIOITS  OF  THE  VARIOUS  P^OJEL  POINTS  AND 

Page 

COICPUTATION  OF  THE  REACTIONS  AT  THE  BASE  OF  EACH  COLUMN-  55 

Page 

!•  Design  of  Bent, Span  Length  40  ft 96 

p Z 

2*  Computation  of  Span  40*-  Col,  Hgt,21*;  Class  I, 

Case  A,  and  Class  II,  Case  A,... ••••••• 


3,  Computation  of  ZJ:  , Span  40*-  Col.Hgt,  21*;  Class  II, 

A 


97 


99 


Case  B,  and  Class  II,  Case  C. 

4.  Reactions  at  Column  Bases  Due  to  a Vertical  Load  of 

1 Lh.  at  Various  Panel  Points;  Span  40 ’-Column 

Height  21*;  Class  I, Hinged  Columns,  and  Class  II, 

Fixed  Columns*  101 

5*  Design  of  Bent, Span  Length  20  Ft* 102 

6*  Design  of  Bent,  Span  Length  30  Ft*  103 

7*  Design  of  Bent,  Span  Length  50  Ft*  * • • 104 

8*  Design  of  Bent,  Span  Length  60  Ft*  • * * * 105 

9*  Reactions  for  Uniformi  Vertical  Load,  of  1 lb*  per  Foot 

Length  of  Truss 106 

10*Reactions  for  Vertical  Load  of  1 Lh*  at  Point  1 * . * . 107 

ll*Reactions  for  Vertical  Load  of  1 Lb*  at  Point  4 * * * * 109 

12*Horizontal  Reactions, Class  I, Case  A,ColT2mns  Hinged; 

Combined  Wind  and  Vertical  Loads  **•••*.*•  Ill 

13*Ratio  Hinged  Columns;Wind  Load  Combined  with  Various 
H 

Vertical  Loads 114 

14*Horizontal  Reactions, Class  II,  Case  A, Columns  Fixed; 

Combined  Wind  and  Vertical  Loads  •**•*..•  115 


IV. 

. TT^  Page 

15.  Ratio  Fixed  Columns;  Wind  Load  Combined  with 
H 

Various  Vertical  Loads 118 

16#  Resisting  Moments,  Class  II,  Case  C;  Combined  Wind  and 

Vertical  Loads  •••••••••  119 

17.  Points  of  Contraflexure , Fixed  Columns;  Wind  Load  20  Lbs. 

per  Square  Foot;  Various  Vertical  Loads  . 122 

18.  Comparison  of  Reactions  Obtained  by  Model  and  by  Theory; 

Span  40»-Col.  Hgt.  21* 127 


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1. 

2 to 

7 to 

21  to 

17  to 

21. 
22.  to 
27. 


LIST  OF  FIGURES 

Line  Diagrams  of  Bent.  40-Ft.  Span  and  21-Ft. 

Coltunn  Height  

6.  Diagrams  to  Accompany  the  Solution  for  the  Horizontal 
Reactions  on  the  Hinged  Columns, Class  I,  Case  A.  . 
10. Diagrams  to  Accompany  the  Solution  for  the  Horizontal 
Reactions  on  the  Fixed  Columns, Class  II, Case  A.  . . 
16. Diagrams  to  Acconpany  the  Solution  for  the  Vertical 
Reactions  on  the  Fixed  Columns, Class  II, Case  B.  . • 
20. Diagrams  to  Accompany  the  Solution  for  Resisting 

Moments  on  the  Fixed  Columns, Class  II,  Case  C . . . 

Line  Diagrams  of  All  Bents  Analyzed 

26. Line  Diagrams  of  One  Bent  of  Each  Span  Length  . . . 

Diagram  of  Paper  Model 


Page 

128 

.128 

132 

136 

139 

142 

143 
145 


■*  f 


VI. 

LIST  uF  GRAPHS 

Page 

!•  Horizontal  Reactions  for  1 lb.  per  foot  span  of  Truss, 

Vertical  Load}  Ratio  - vs«  Column  Height.  146 

2*  Horizontal  Reactions  for  1 lb.  per  foot  Span  of  Truss, 

Vertical  Load;  Ratio-  vs.  Span  Length  147 

3.  Resisting  Moments  for  1 lb.  per  foot  Span  of  Truss, 

Vertical  Load;  Ratio-  ^ vs. Column  Height  .......  148 

4.  Resisting  Moments  for  1 lb.  per  foot  Span  of  Truss, 

M 

Vertical  Load;  Ratio  - vs.  Span  Length  149 

6.  Vertical  Reactions  for  1 lb.  Load  at  Point  4;  Right-Hand 

Column, Vertical  Load;  Ratio  - Colupi  Height  . . 150 

6.  Vertical  Reactions  for  1 lb.  Load  at  Point  4;  Right-Hand 

Column,  Vertical  Load;  Ratio-  vs.  Span  Length  • • . 151 

t 

7o  Vertical  Load  30  lbs.  per  square  foot, Wind  Load  20  lbs. 

per  sq,  ft.;  Ratio  - ^ vs. Column  Height.  ...  ...  152 

g^Vertical  Load  45  lbs.  per  sq.  ft;  Wind  Load  20  lbs. per  sq. 

ft.;  Ratio  - vs.  Column  Height 153 

H 

9.  Vertical  Load  60  lbs.  per  sq.  ft;  Wind  Load  20  lbs.  per 

sq.  ft;  Ratio  - vs.  Column  Height  .........  154 

10.  Vertical  Load  75  lbs. per  sq.ft.; Wind  Load  20  lbs. per  sq. 

ft. ; Ratio  - S^s.  Column  Height  . 155 

11.  Vertical  Load  100  lbs.  per  sq.  ft.;  Wind  Load  20  lbs.  per 

sq.  ft.;  Ratio  - Hr  vs.  Column  Height 156 

H 

12.  Vertical  Load  30  lbs.  per  sq.ft;  Wind  Load  20  lbs. per  sq. 

ft.;  Ratio  - & vs.  Span  Length  157 

13.  Vertical  Load  45  lbs.  per  sq.  ft.;  Wind  Load  20  lbs.  per 

sq.ft;  Ratio  - §R  ys.  Span  Length 158 

H 

14.  Vertical  Load  60  lbs.  per  sq.  ft.;  Wind  Load  20  lbs.  per 

sq.ft.;  Ratio  vs.  Span  Length 159 

H 


VII. 

15.  Vertical  Load  75  lbs.  per  sq..  ft.;  Wind  Load  'dO  lbs.  per 

Sq.ft.;  Ratio  - SR  vs.  Span  Length 160 

H 

16.  Vertical  Load  100  lbs.  per  sq.  ft;  Wind  Load  20  lbs.  per 

S^.  ft. ; Ratio  vs.  Span  Length 161 

H 

17.  Vertical  Load  30  lbs.  per  sq..  ft.^.  Wind  Load  20  lbs.  per 

sq.  ft.;  Ratio  - |R  ts.  Ratio  - 162 

18.  Vertical  Load  45  lbs.  per  sq.  ft..  Wind  Load  20  lbs.  per 

sq.  ft;  Ratio  - ^ vs.  Ratio  - 

19.  Vertical  Load  60  Lbs.  per  sq.  ft..  Wind  Load  20  lbs.  per 

sq.  ft.;  Ratio  - Hpvs.  Ratio  - Colnmn  Height 164 

H Span  Length 

20.  Vertical  Load  75  lbs.  per  sq.  ft.;  Wind  Load  20  lbs.  per 

sq.  ft.;  Ratio  - |R  vs.  Ratio  - 1^5 

H Span  Length 

21.  Vertical  Load  100  lbs.  per  sq.  ft. ^ Wind  Load  20  lbs.  per 

sq.  ft.;  Ratio  - vs.  Ratio-  166 

H Span  Length 

22.  Points  of  Contraf 1 exare ; Vertical  Load  30  lbs.  per  sq.  ft.. 

Wind  Load  20  lbs.  per  sq.  ft.;  Ratio-X  vs  Col.  Heightl67 

d 

23.  Points  of  Contraflezure; Vertical  Load  45  lbs. per  sq.ft.. 

Wind  Load  20  lbs. per  sq.ft. ; Ratio-^  vs.  Col.  Height  .168 

d 

24.  Points  of  Con traflexnre; Vertical  Load  60  lbs. per  sq.ft. 

Wind  Load  20  lbs. per  sq.ft. ;Ratio-Y  vs.  Col. Height..  169 

d 

25.  Points  of  Contraflextu’e;  Vertical  Load  75  lbs. per  sq.ft.. 

Wind  Load  20  lbs. per  sq.ft.;  Ratio-^  vs.  Col.  Height  170 

26.  Points  of  Contraflexiire;  Vertical  Load  100  lbs.  per  sq.ft.. 

Wind  Load  20  lbs.  per  sq.ft.;  Ratio-X  vs.  Col. Height  171 

27.  Points  of  Contraf lexure ; Vertical  Load  30  lbs.  per  sq.ft.. 

Wind  Load  20  lbs.  per  sq.ft.;  Ratio-i  vs.  Span  Length  172 

28.  Points  of  Contraf lexure;  Vertical  Load  46  lbs.  per  sq.ft.. 

Wind  Load  20  lbs.  per  sq.ft.;  Ratio-Z  vs.  Span  Length  173 

d 


VIII 


29*  Points  of  Contraflexure;  Vertical  Load  60  ITds,  per  sq.ft., 

Wind  Load  20  Tos.  per  sq.ft;Ratio-  ^ vs.  Span  Length... 174 

30.  Points  of  Contraflexure;  Vertical  Load  75  Ihs.per  sq.ft.^  j 

Wind  Load  20  Ihs.  per  sq.ft;  Ratio-^  vs.  Span  Length... 175 

31.  Points  of  Contraflexure;  Vertical  Load  100  Ihs.per  sq.ft.^ 

Wind  Load  20  Ihs.  per  sq.  ft;  Ratio-^  vs. Span  Length... 176 

32.  Points  of  Contraflexure; Vertical  Load  30  Ihs.per  sq.ft.,  | 

Wind  Load  20  Ihs.per  sq.ft;  Ratio  - j vs.  Ratio  - ! 

Column  Height  -irjrj 

Span  Length * •••••  | 

33.  Points  of  Contraflexure; Vertical  Load  45  Ihs.per  sq.ft.,  | 

Wind  Load  20  Ihs.per  sq.ft;  Ratio-J  vs . Rati o 178  i 

d Span  Length  j 

34.  Points  of  Contraflexure; Vertical  Load  60  Ihs.per  sq.ft.^ 

Wind  Load  20  Ihs.per  sq.f t;Ratio-J  vs. Ratio 179 

* * d Span  Length 

35.  Points  of  Contraflexure; Vertical  Load  75  Ihs.per  sq.ft.,  | 

Wind  Load  20  Ihs.per  sq.ft;  Rat  io  vs  .Ratio  -"|;^'an^§en§t'h  ' 

36.  Points  of  Contraflexure;  Vertical  Load  IQO  Ihs.per  sq.ft.,  j 

„ _ _ Y _ Col. Height  i 

Wind  Load  20  Ihs.per  sq.ft;  Rat  io--^  vs  • Ratio  -span  Length 


IX 


TABLE  OF  SYMBOLS, 

A - area  of  section  of  member. 

C - unknown  horizontal  force  at  the  base  of  either  column  when 
forces  or  moments  are  applied  at  the  base  of  the  right-hand 
column . 

d^-  deflection  at  any  point  x of  the  truss  or  col'jmn  in  any  direc- 
tion. 

d - distance  from  base  of  column  to  foot  of  knee  brace. 

E - modulus  of  elasticity. 

F - unknown  force  developed  at  the  top  of  either  column  when  forces 
or  moments  are  applied  at  the  base  of  the  right-hand  column. 

H - any  horizontal  reaction;  also  the  total  horizontal  component  of 
the  wind  forces  acting  on  the  bent. 

- horizontal  reaction  on  right-hand  column. 

Hl  - horizontal  reaction  on  left-hand  column, 

h - height  of  column. 

I - moment  of  inertia  of  column  section. 

k - lever  arm  of  force  which  produces  unit  moment  at  the  base  of  the 
right-hand  column. 

'L  - length  of  any  truss  member. 

L - length  of  span  of  bent. 

M - resisting  moment  at  the  base  of  either  column. 

Mp_  - resisting  moment  at  the  base  of  the  right-hand  column. 

- resisting  moment  at  the  base  of  the  left-hand  column. 

N - either  the  horizontal  or  vertical  reaction. 

P - stress  in  any  truss  member  due  to  various  conditions  of  loading. 

P-j^  - stress  in  any  trues  member  due  to  a horizontal  force  of  one 

pound  applied  at  the  base  of  the  right-hand  column,  when  the 


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columns  are  hinged. 

Pg-  stress  in  any  truss  member  due  to  a horizontal  force  of  one 
pound  applied  at  the  base  of  the  right-hand  column,  when  the 
columns  are  fixed. 

Pg-  stress  in  any  truss  member  due  to  a vertical  force  of  one  pound 
applied  at  the  base  of  the  right-hand  column,  when  the  columns 
are  fixed. 

P^-  stress  in  any  truss  member  due  to  a moment  of  one  inch  pound 
produced  at  the  base  of  the  right-hand  column  by  a force  ap- 
plied at  the  end  of  the  lever  BK. 

Q - numerical  term  in  the  expression  for  the  value  of  Pg. 

R - the  unknown  vertical  force  acting  at  the  base  of  either  column 
when  a unit  moment  is  applied  at  the  base  of  the  right-hand 
column. 

S - coefficient  of  F in  the  expression  for  the  value  of  Pg. 

t - total  height  of  bent. 

T - coefficient  of  R in  the  expression  for  the  value  of  P^. 

U - the  stress  in  any  truss  member  due  to  a unit  force  applied  at 
any  point  where  the  deflection  in  some  direction  is  desired. 

V - vertical  reaction  at  the  base  of  either  column. 

- vertical  reaction  at  the  base  of  the  right-hand  column. 

- vertical  reaction  at  the  base  of  the  left-hand  column. 

W - total  load  at  any  panel  point  of  the  trues. 

w - wind  load  in  pounds  per  linear  foot  on  the  windward  column. 

Y - distance  from  base  of  column  to  point  of  contraf lexure  of  eithea 

column. 

Yj^  - distance  from  base  of  right-hand  col^jmn  to  point  of  contra- 


flexure. 


I 


XI 


Yl  - distance  from  base  of  left-hand  column  to  point  of  contra- 
flexure  . 

- deflection  of  any  point  on  the  elastic  curve  of  either  column 
away  from  the  tangent  at  some  other  point. 

^JL  - deformation  of  the  bottom  chord  of  the  truss. 


1. 


I.  INTRODUCTION 

1.  Preliminary,-  A mill  building  bent  of  the  ty^pe 
shown  in  Pig,l  is  a statically  indeterminate  structure.  The 
deflection  of  the  truss  under  vertical  loads  effects  bending  in 
the  columns  and  develops  forces  at  the  points  where  the  truss  is 
connected  to  the  columns.  These  forces  cause  stresses  in  the 
knee  braces  and  in  the  members  of  the  truss  which  are  statically 
indeterminate. 

There  are  two  classes  of  this  type  of  structure  herein 
considered.  The  first  is  ClassI,  columns  hinged  at  the  base. 

For  either  vertical  or  wind  loads  there  are  four  unknown  reactionb 
the  horizontal  and  vertical  reactions  at  the  base  of  each  column. 
The  three  equations  of  static  equilibrium  are  not  sufficient 
to  determine  these  reactions,  and  an  additional  equation  must  be 
obtained.  The  second  class  is  ClassII,  columns  fixed  at  the  base. 
Here  there  are  six  unknown  reactions,  the  vertical  reaction,  the 
horizontal  reaction,  and  the  resisting  moment  at  the  base  of  each 
column.  The  three  equations  of  static  equilibrium  must  be  aug- 
mented by  three  other  equations  to  determine  these  unknown 
reactions. 

In  designing  practice  the  effect  of  the  deformation 
of  the  truss  under  vertical  loads  on  reactions  and  stresses  is 
neglected  in  both  classes  given  above.  The  stresses  in  the  raent- 
bers  of  the  truss  due  to  vertical  loads  are  then  the  same  as  if 
the  truss  were  supported  on  solid  walls.  The  columns  are  designed 
to  resist  bending  due  only  to  wind  forces,  and  to  resist  direct 
compression  due  to  both  wind  and  vertical  forces.  The  knee  braces 


2 


are  designed  for  the  stresses  due  only  to  wind  loads. 

In  the  case  of  hinged  columns,  the  fourth  equation 
necessary  to  determine  the  reactions  due  to  wind  loads,  is 

I 

obtained  in  practice  by  the  assumption  that  the  horizontal  reac-  | 
tions  are  equal.  The  three  additional  equations  necessary  in  | 

I 

the  case  of  fixed  columns  are  obtained  by  assuming  that  the  j 

horizontal  reactions  are  equal  and  by  assuming  the  location  of  j 
the  point  of  contraflexure  in  each  column.  | 

I 

I 

The  point  of  contraflexure  is  usually  arbitrarily  i 

j 

assumed  as  midway  between  the  foot  of  the  Imee  brace  and  the  base 
of  the  column,  it  i-s  evident  that  the  columns  are  not 

rigidly  fixed  at  their  bases,  the  point  of  contraflexure  is  often 
assumed  to  be  located  at  a point  a distance  above  the  base  i 

equal  to  one-third  to  one-fourth  of  the  distance  from  the  base  I 

of  the  column  to  the  foot  of  the  ioiee-brace.  Sometimes  the  j 

location  of  the  point  of  contraflexure  is  obtained  from  formulas  | 
based  on  other  assumptions.  Three  of  these  methods  are  as  follows t 
1.  Each  column  is  assuned  to  be  fixed  at  the  base  and  to 

be  hinged  at  the  top,  and  the  horizontal  deflections  at  the  ! 

i 

foot  of  the  Imee  brace  and  at  the  top  of  the  column  are  assumed 
to  be  equal.*  The  location  of  the  point  of  contraflexure  is  then 
found  from  the  equation  Y = ^ x where  Y is  the  distance 


*See  "The  Design  of  Steel  Mill  Buildings”  by  Milo  S 
Ketchum,  Third  Edition,  Pages  87  to  91. 


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3. 

from  the  base  of  the  coltnnn  to  the  point  of  contraflexnre,  d 
is  the  distance  from  the  base  of  the  column  to  the  foot  of  the 
knee  brace,  and  h is  the  height  of  the  column* * 

E.  Each  column  is  assumed  to  be  fixed  at  the  base,  and  the 
portion  of  the  column  from  the  foot  of  the  iaiee  brace  to  the 
top  of  the  column  is  assuned  to  remain  a straight  vertical  line.* 
In  accordance  with  these  assumptions,  the  point  of  oontraflexure 
is  midway  between  the  foot  of  the  knee  brace  and  the  base  of  the 
column* 

3.  Bach  column  is  assumed  to  be  rigidly  fixed  at  its  base 
and  top**.  In  accordance  with  this  assumption  , the  point  of 
contraflexure  is  located  midway  between  the  base  and  the  top  of 
the  column*  This  method  places  the  point  of  contraflexure 
higher  than  the  other  methods* 

A recent  investigation***  has  shown  that,  for  the  type 
of  bent  given  in  Fig.l,  the  ass  trap tion^ that  the  horizontal 
reactions  due  to  wind  loads  alone  are  equal^ is  materially  in 
error.  This  investigation  has  also  shown  that,  while  the  common 
assumption  regarding  the  location  of  the  point  of  contraflexure  is 
not  materially  in  error,  the  bending  moments  in  the  columns  ob- 
tained from  the  assumption  of  equal  horizontal  reactions  are 
materially  in  error* 

'^See  "The  Design  of  Steel  Mill  Buildings"  by  Milo  S.Ketchum 
Third  Edition, pages  87  to  91* 

*5fcSee  "Influence  Lines  for  Bridges  and  Roofs"  by  Burr  and  Falk, 

Third  Edition,  Page  53* 

***See  Thesis, "Analysis  of  Statically  Indeterminate  Building 

Trusses  by  the  Application  of  Maxwell* sTheorm  of  Reciprocal  Bis- 
placements"by  S.R.Offutt, University  of  Illinois,  1980* 


4. 


The  writer  has  been  imable  to  find  any  other  investi- 
gation of  the  effect  of  the  deformation  of  the  truss  under 
vertical  loads  upon  the  reactions  at  the  column  bases. 

2.  Scope  and  Purpose.-  The  method  of  solution  pre- 
sented in  this  thesis  may  be  used  to  find  the  reactions  at  the 
bases  of  the  columns  of  any  structure  , consisting  of  a truss 
of  any  type  supported  on  two  columns,  and  connected  to  each 
column  at  the  top  of  the  column  and  at  some  other  point  by  a 
knee  brace. 

The  quarter-pitch  Pink  truss  supported  on  two  columns 
is  a common  type  of  structure  used  in  mill  buildings,  and 
has  been  chosen  for  this  investigation.  The  writer  hoped  to 
present  empirical  curves,  from  which  the  reactions  at  the  column 
bases  of  any  bent  of  this  type  due  to  combined  wind  loads  and 
vertical  loads  might  be  obtained,  but  he  found  that^ because  the 
reactions  due  to  vertical  loads  were  dependent  upon  so  many 
characteristics  of  the  bent,  this  was  impossible  with  the  limited 
data  available. 

The  purpose  of  this  investigation  is  to  study  the 
effect  of  variations  in  different  characteristics  of  the  bent 
upon  the  reactions  due  to  vertical  loadSj  and  to  determine  whether 
or  not  the  assumptions  regarding  horizontal  reactions  and  the 
location  of  the  points  of  contrafleocure  are  materially  in  error^ 
when  the  bent  is  subject  to  both  vertical  and  wind  loads. 

3.  Acknowledgements . - This  investigation  was  made  under 
the  supervision  of  Professor  W.  M.  Wilson,  Research  Professor 

of  Structural  Engineering  at  the  University  of  Illinois,  who  gave 


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6 


the  writer  many  helpful  suggestions. 

The  method  used  in  the  analysis  of  the  bents  was 
formulated  by  C.  A.  Ellis,  formerly  Professor  of  Structural 
Engineering  at  the  University  of  Illinois.  The  values  of  the 
reactions  due  to  wind  loads  were  taken  from  a thesis  written 
by  Mr.  S.  R.  Offutt*.  The  vertical  deflections  of  the  various 
panel  points  of  the  bents  analyzed  were  scaled  from  the 
Williot  diagrams  prepared  by  Mr.  Offutt  for  his  thesis.  The 
writer  also  used  sane  of  his  figures  and  tables. 

The  writer  wishes  also  to  acknowledge  his  indebted- 

I 

ness  to  the  Department  of  Theoretical  and  Applied  Mechanics  for  j 

1 

the  use  of  the  microscope  and  micrometers  belonging  to  the 
Department,  which  he  used  in  his  test  of  a paper  model  of  a 
bent. 


*See  footnote  on  page  3. 


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6 


II.  OUTIOE  OF  METHOD 

4.  General  Application  of  Maxwell*  3 Theorem  of 
Reciprocal  Displacements.-  Maxwell*s  theorem  of  reciprocal 
displacements  establishes  a mutual  relation  betw^een  any  two 
points  of  a structure.  This  theor«?/»,  which  may  easily  be 
proved,  may  be  stated  as  follows: 

The  displacement  of  any  point  A of  a structure  in  any 
direction,  as  horizontal,  due  to  a force  P applied  at  any  other 
point  B of  the  structure  in  any  direction,  as  vertical,  is 
equal  to  the  displacement  of  point  B in  the  vertical  direction 
due  to  the  force  P acting  in  a horizontal  direction  at  A.* 

This  principle  may  be  easily  applied  to  determine  the 
reactions  of  an  indeterminate  structure,  such  as  a mill  building  j 
bent  of  the  tupe  shown  in  Fig.l.  Suppose  it  is  desired  to  find  j 

I 

the  horizontal  reaction  at  point  B due  to  a vertical  load  of  one  | 
pound  at  point  3 (See  Fig.3) • The  method  is  as  follows:  | 

Apply  a horizontal  force  of  one  pound  at  point  B ahd 
find  d3,  the  horizontal  displacement  of  point  B,  and  d0,  the  vertL 
cal  deflection  of  point  3.  By  Maxwell* s theorem,  dg  equals 
the  horizontal  displacement  of  point  B due  to  a vertical  load 
of  one  pound  at  point  3.  If  dg  equals  the  horizontal  displace- 
ment of  point  B due  to  a vertical  load  of  one  pound  at  point  3 
and  dg  equals  the  horixontal  displacement  of  point  B due  to  a 
horizontal  force  of  one  pound  at  B,  then  the  horizontal  re- 
action at  B due  to  a vertical  load  of  one  pound  at  3 is 

HB  = II  ^ 1 It. 

^Porfl complete  statement  of~this  theorem  see  Militor*s  "Xinetic 
Theory  of  Engineering  Structures”  (1910  edition)  pages  27-29. 


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7 


Thus,  the  deflection  diagram  of  the  structure  when 
subject  to  a horizontal  force  of  one  pound  at  B is  the  influence 
line  for  the  horizontal  reaction  at  B* 

Similarly^ it  may  be  shown  that  the  deflection  diagram 
of  the  structure  when  subject  to  a vertical  force  of  one 
pound  at  a reaction  point,  is  the  influence  line  for  the  verti- 
cal reaction  at  that  reaction  point.  j 

Similarly,  I«Iaxwell*s  theorem  may  be  applied  to  finding 
the  resisting  moment  at  any  reaction  point.  For  example, 

I 

suppose  it  is  desired  to  find  the  resisting  moment  at  point  B oi 
the  structure  given  above  due  to  a vertical  load  of  one  pound  at  | 

I 

I 

point  3 (See  Fig.l9).  Proceed  as  follows;  | 

i 

Apply  a force  of  l/k  pounds  at  the  end  of  a perfectly  | 

rigid  lever  arm  M of  length  K inches^  rigidly  attached  to  the 

column  at  B.  The  resulting  moment  at  B is  one  inch  pound. 

Find  d]£  + 4,  the  horizontal  displacement  at  the  end  of  the 

lever  arm  due  to  the  moment  of  one  inch  pound  at  B.  Find  d„, 

o 

the  vertical  deflection  of  point  3 due  to  the  moment  of  one  ' 

inch  pount  at  B^or  due  to  the  force  of  l/k  pounds  acting  at  X. 

By  Maxwells  theorem,  dg  equAls  what  the  horizontal  displacement 
at  the  end  of  the  lever^ would  be  due  to  a force  of  l/k  pounds 
acting  vertically  at  point  3,  or  equals  l/k  times  what  the 
displacement  would  be  at  X due  to  a vertical  load  of  one  pound 
acting  at  point  3.  Then  the  horizontal  displacement  at  the  end 
of  the  lever  X due  to  a force  of  one  pound  acting  vertically  at 
point  3 equals  kdg*  Since  kd^  is  the  horizontal  displacement  at 

at" 

the  end  of  the  lever  X due  to  a vertical  force  of  one  pornl  acting 


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-04  o,T'  ■'. 

0 ootc 

X i .viJlV 

Y - 

O f 0 ■•■■  T 

•^0." 

-j.':] 

:-c,  u-t'i.u.n  ; 

8 


point  3,  and  d^^  + is  the  horizontal  displacement  at  the 
end  of  the  lever  due  to  a moment  of  one  inch  pound  at  B,  then 
the  moment  at  B due  to  a vertical  force  of  one  pound  at  point  3 
is 

k da 

M-r=  ^ xl  inch  potind. 
dK  + 4 4 ^ 

Thus^the  deflection  diagram  of  the  structure  when 
subject  to  a moment  of  one  inch  pound  at  a reaction  point  is  the 
influence  diagram  for  the  resisting  moment  at  that  reaction 
point. 

This  principle  will  now  be  used  to  determine  the 
reactions  for  a typical  mill  building  bent  when  subject  to 
vertical  loads.  ^ line  diagram  of  the  structure  is  shown  in 
Fig.l.  Table  1 gives  the  design  and  general  data. 

There  are  two  classes  to  be  considered,  viz: 

Class  I,  columns  hinged  at  their  bases,  and  Class  II,  columns 
fLxed  at  their  bases.  These  two  classes  will  be  considered 
in  the  order  named. 

5.  Application  to  Class  1,  Columns  Hinged  at  Their 
Bases."  In  this  case  there  are  no  resisting  moments  at  the 
bases  of  the  columns^ since  the  columns  are  free  to  turn  at 
their  bases.  The  vertical  reactions  may  be  determined  by  the 
equations  of  static  equilibrium.  Thus,  only  the  horizontal  re- 
actions are  statically  indeterminate. 

Case  A.  Horizontal  Reactions.-  The  influence  line 
for  the  horizontal  reaction  at  B,  which  is  the  deflection  diagram 
of  the  structure  when  it  is  subject  to  a horizontal  load  of  one 
pound  at  B,  will  now  be  determined.  Pig#  3 shows  this  deflection 
diagram  greatly  exaggerated. 


9 


Apply  a horizontal  force  of  one  poimd  at  B (Pig* 3). 

An  equal  and  opposite  force  must  he  applied  at  B*  These  forces 
cause  bending  in  the  columns  and  produce  horizontal  forces  at 
the  points  where  the  columns  are  connected  to  the  truss*  These 
forces  acting  on  the  columns  are  statically  determinate  (See  Pig. 
8)  and,  when  referred  to  the  truss,  are  equal  and  opposite* 

Consider  the  point  D to  be  fixed  in  position  and  the 
point  B (Pig* 3)  to  be  free  to  move  to  B£,  due  to  the  bending  of 
the  columns  and  the  defoliation  of  the  truss*  Both  columns 
are  free  to  rotate  at  their  bases*  The  solid  outline  (Pig*3) 
shows  the  initial  position  of  the  structure*  The  broken 
outline  shows  the  structure  with  its  i nitial  form,  but  dis- 
placed through  the  horizontal  distance  DDi=BBi  due  to  the  bend- 
ing of  the  column  BG*  The  dotted  outline  shows  the  structure 
displaced  to  its  final  position  due  to  the  deformation  of  the 
truss*  Point  B has  moved  from  B to  Bq_  due  to  the  flexure  of  the 
column  BG,  and  from  B^^  to  its  final  position  B£  due  to  the 
deformation  of  the  truss  and  the  flexure  of  the  column  BK*  The 
deflection  diagram  can  not  be  drawn  to  scale  until  the  flexure 
of  the  columns  and  the  deformation  of  the  truss  have  been 
determined*  First,  the  flexure  of  the  columns  will  be  detemined. 

Flexure  of  the  Columns: - 

Since  the  forces  acting  on  the  two  columns  are 
equal,  but  opposite,  the  flexure  of  the  columns  will  be  the 
same  in  magnitude,  but  opposite  in  direction*  The  oolumn  acts 
as  a beam  under  the  horizontal  forces.  BGiJg  (Pig* 4)  shov/s  the 
elastic  curve  of  the  deflected  column  BGJ*  Braw  a tangent  to 


- V.  - ' ' 


7 

«>  « 


■;;^  3-1  .'■•.^jic‘:  .•;rTO'J  t' ' ';dOq[c|' 

‘ii 

’i'X'C'fi  aviJ^O’i-.  1 ■ ■^:."  iii  n.;  I £»i  ;<,•*<-' 


'j  -j  • 


I o . . : • j 


■ ' 0'-‘  n.: 


O v 

; OGtOU.Ti'fC'J  1 : ■ 

■•  fi/lTJCrj-GG 

ant  o v 

TO.;  ,>'i> 

h'’  : - gxx  • ' ^ - 

: :sr'ij  too 

ry  ' 4*./V 

■ - =■ 
y ^ 

■J 

OT  , . n 

‘ Oi-iXJv  0 1 

noTTfil'ita' 

r . 

.'  •*•-:■  oii  Ov^  ( 

I t lit  „<> 

lit  Zr.-.'i, 

r ‘ 

;rX  , • ' . / ) /v.;: 

ct  0 '.‘  VI 

OO  ' 

J « 

oX'^..L':c'JO-ft  fif"  .h 

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■f'  tiv..; 

4.  - V -V 

i,  -Vi  .-' 

-'•tif' 

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h fif? 

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'"'"cdn 


1 


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DiU  •'to  ;:oj;r  , .to'' ' ^ o:;  o;':6  tv  j-.;  X ::ocv  "j  K ? J vf^O'iX  . ■ K' 

rf*'  !».o  OT -rr-  X?  t?.' ‘ of  o.*r*  j- ■ o ' il  i'C.  i^.vv  ; n a 

•-  • •’■  •''?■  • ' iiiri , .tiioj  X'  X.;;  .V.:  0./  j-/'  . ot";  , ry,  •:  ,;.:or- 

» " T:-:  0T;rx©X';  r - ' o:’’.  .■  ..  ' '''••/■.’■  ... 

’’k'  ‘ 1 

• an?  IX  v>;  ..  .'. 'oa  r‘  t?  -iX.  oo  .ton  r;;iv  .-•  rtv '.;  s.  if  ■ ‘ 


k'r  ( 


n..a/  ov  • 

ou  ■_  rcoX  ofi  ao  * 


.!c  ii  : ' 'r-^OJi^h  ant  I, 


'* 


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■>  r. 


6.^  *1*1.1 
'i.,t  3t  :‘iX  ' . ' 

t 

c^^vj  : n y:£r  o '-  . 


■ ,V^-s.r  roo"‘•^  _ 

i ,:  tt  ' 

o-X'-'Kirf  6:  ” 

- : p.rj.  .■  ,'OL*  i>:‘t  '1'  D-.l-'TOri 

,.  X no  :'  .Xv'O..  v.f)£>TO  . C4'’  'X-  ■'"■ 

( 

■■J  • . . a iu  * ...  4.  G - ■ • .'w  .-^vi  a«j  . • o f ». 


.,'t  ar;c--'‘ 


ut  t;-!>  Vi'l£.f  •;  •*'. 


*■;.#  .-.i>,yTC>  I )tnOoXiO.:X  i>ilt  TV 

-.■rrTiwr  '■?  g-‘- ' . 


10 


the  elastic  curve  of  the  column  at  point  and  produce  itto 
intersect  the  horizontals  through  D and  J*  Then  Ao  e<itials 

I 

I 

the  horizontal  displacement  of  point  Jg  away  from  the  tangent  i 
to  the  elastic  curve  at  G^,  and  equals  the  displacement  of  l 
D away  from  the  same  tangent*  Both  A and 
found  by  the  area-moment -Tnethod. 

The  area-moment  method  states  that  the  deflection  of  | 

any  point  on  a beam  subj'ected  to  flexure  away  from  the  tangent  I 

I 

to  the  elastic  curve  at  any  other  point  of  the  beam,  is  equal  to  j 
the  statical  moment  of  the  area  of  the  M/EI  diagram  included  | 

between  the  two  points,  about  the  point  where  the  deflection  is 
desired*  The  bending  moment  diagram  of  the  column  is  dgj  (Fig* 4).  | 


A - 180  X 90  X 120  1,944,000  23.940.89,  . 

- ^ *=  — ^ ‘inches. 


El 

180  X 36  X 48 


311jp40 


3,830.54  . - 

inches. 


hand 


“ SI  ^ El 

Before  the  horizontal  displacements  of  the  left^column 
at  points  J and  G,  dj-  and  d(j  respectively,  can  be  found,  dqi^ 
the  horizontal  displacement  of  J due  to  the  deformation  of  the 
truss,  must  be  found*  This  will  now  be  done* 

Deformation  of  the  Truss’, - 

The  forces  acting  on  the  truss  are  shovm  in  Fig.2* 

The  horizontal  deflection  of  point  J due  to  the  deformation  of  th( 
truss  may  be  found  either  algebraicly  or  graphically 

Algebraic  Method:  dx<=  ^ * 


"^Por  the  history  and  development  of  this  formula  see  article 
by  G.P. Swain  in  Jour,  of  Franklin  Institute,  Vol*85 , page  102, '*0n 
the  Application  of  the  Principle  of  Virtual  Velocities  to  the 
Determination  of  the  Deflection  and  Stresses  of  Frames”. 


,V  ,'/■ 


■x  J-  t 


N ■ ^ r. 


4O  t'TN 

■ \ • - ■ 

«V">r 


-■  I...  ti,  ■>■ 
I ■.<if'r''  - ■ V' 


di 


yi  !). 


■'Ti  trcl0:s 


jt 


.:  C s_ 


I’l'^ 


. Jiv  il 


* ' I r 
>.  '■•  ■*  • 


. ‘J 


,) 


v» 


> * l-l  . 

VTM  U 


j:  i v-  j 


/ V 


• A ^ 

♦i»  *1'"*’  *••  • • -'s  ■i 


0 


-U’ 


•1 


io 


Li.'JJf 


1 


) 


-'.  «•  • .4 


4'; 

• ■ V 


x;  ■ 


o 


's^:  'to  Q 


1:: . * .'  os. .!.  y K 'i  : 

.1  i*  ’j:.'.'  magft 


11. 

Where  ^ is  the  displacement  of  any  panel  point  x of  the  truss  in 

di/e. 

any  direction  in  inohes^to  the  given  loads. 

P is  the  stress  in  any  member  in  pounds  due  to  the  given 
loading. 


U is  the  stress  in  any  member  in  pounds  due  to  a unit  load 
acting  at  the  point  where  the  deflection  is  measured  and  in  the 
direction  of  the  measured  deflection. 

2^  is  the  length  of  any  member  in  inches. 

A is  the  sectional  area  of  any  member  in  square  inches. 

E is  the  modulus  of  elasticity  in  pounds  per  square  inch,  and 
is  constant  for  all  members  when  they  are  made  of  the  same  material* 
P and  U are  positive  or  negative  according  as  the  stress  is  tension 
or  compression.  The  quantityPU  ^/a  is  found  for  each  member,  and 

the  algebraic  sum  of  these  quantities,  when  divided  by  E, gives  the 
desired  deflection.  The  horizontal  deflection  of  point  J relative 
to  point  Gr(Pig.S)  .dll.  due  to  the  deformation  of  the  truss  is  de- 


sired. 


^11=  z ^ 


a.  is  the  stress  in  any  member  due  to  the  forces  shown  in 


Pig. 2. 


la  is  the  stress  in  any  member  due  to  a force  of  one  pound 
acting  horizontally  at  J(Pig*5).  To  keep  the  truss  from  moving  to 
the  right  under  the  unit  force  at  an  equal  and  oioposite  force 
must  be  applied  at  G(Pig.5) .Since  the  horizontal  displacement  of  J, 
due  only  to  the  deformation  of  the  truss  and  not  to  the  rotation  of 
the  truss  about  G,  is  desired,  an  upward  vertical  force  must  be 
applied  at  K and  an  equal  downward  force  at  G to  form  a couple  whose 
moment  is  equal  and  opposite  to  the  moment  of  the  couple  due  to  the 

unit  horizontal  forces  at  G andJ.  This  couple  prevents  rotation  of 
the  truss  about  G.  The  magnitude  of  each  vertical  force  equals  Itimes 


. V ' 


• t; 


on:r  'xu  ?.  X*.»j:fciq  It-:  l.  • . 3 o- < Lyfi 

« - 

'TX' ‘ ... ' c.)  ^ ^;«j(ic^.U  ni  lie-:  tX  Xitr. 

.fi  Lvnor^s' 


OC..I  OJ 


e£r(\ 


f 


f 


.'  oX  t.*  •'.  . 'iicjoq  nL  'jra.-L'cr/r  *i;rj'.v  £x.f  seQ'j . r;  *fi  ^2  Q 

0;  ■ L'i  ijita  r • 


•;  u or{;t  2ryi<>q^g^^  ‘ : 3f/i ?•.•:  / j 

, '.’’V  <' 

. i . t 'W  ;.L  t:  h\'.  i-  :fl*37&X ‘W...  :;: 


a'T  i.' 


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fT^.  J 


c ■ Jl;: ' ^ 1'  ,' 

■- •■’  ■:  ' , i-:  U Ti-^oitv  Xo_  -TO  ■.Jl'j;..n:.  O'"  <ej  i: 

' ■•  o^h.r.,  t3T..i  '"i'r.;-  ",  .'u'  v.;-:wa;a-'4ii  i..'  ■X'.''r  tii.'tHCOO  r; 

I .V. 

X;  '>'ro.or.'v/  o.* ..^opi  f-fir-rui  o:: . O'  v-n-s 

. 'V*.;  ..•,>.-t>  To'i:  Jb.l»T.>'j.  <-l2  . iJ  iv;:  i.' ;:  .7  t>'iC^u;06  *XQ 


I 


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, 'j'‘/-.'.  7;>  o.'.oxit  'i  ;)i- T ; O'.' 


r ► 


. , ‘j  j -iOv?  !)  j i .1  X 1.  . j.  . 0^.'U  '.^ii’j.* 

i.  ..  tf'j:.  V ‘10  } L . ' :>.f  tt.: 


O*.'.  J.  ' Or>  A>S>7i.,';0L'  ,|. 


. ;fx^2■J<'x  0;  I; 


',.  1:;  ..4  V :'r.,  ::i'r^  ^fi  j;^ 


1 


».• 


X.. o.r^'"  aO  cj'  »ir,&  'X>'T  ...  \:x:'  cil  e c orf^  £’ ^ '‘if 

oj-  ?;jjkX7c  • iiS'/'ii  1'.';  q003i  o-i*  . .’X  17,.'^.  XX.ri 2*:^.  •i-.:orl 


■:0 


0": 

C’ X XwO-' , c f>;i 

X .;m,  iX:-  ^ 

' * s 

i'-oX  JiL:.'  O ' 

to'lj.f''''  trt!^  *■'1  ,: 

••  *^’  ir 

OO-  j;  j.i»  1. 

-'  ..  ; :I  ;w;r  ^.. 'X 

01.T  , { ..  .XS).^  X 

^ Dcf  X : ». 

^r. 

/ 

X f>.i'  c.'  .r:. 

7 X'  ir-.xri'l' 

t:  i .'■  X C'  n 0 X X r.?!  0 ;> 

0 ' C'  YlilO  OtX' 

Of."  i 7 

...  X' 

J . 'i  ■ 7 'V'  •.-'sj.u 

, ./'.‘iXbU  r.j i , ' 

- atiJTi;.*'  odt 

’.  ' '-'J  : ‘ 

i'!".  - . O.r 

' ■•  v''I. 

a-:'‘''.’ir7.'o0  .fir  -•. 

) 

*)-Cs  .ii  j -J  *y  X ,k  C^»J* 

‘1  c:‘ 

O.C^  -T  ■ » 

■i?  f'j  ail-  no  i,  . 

I ::_  X i,l  f''/',rrr 

r:  :XO.ct  X 

0'2  q? 

oXqiJC  5 ,'f 

, j.i '/  tt 

* f 

X i ' .' , . J*- .trji ,- 

'f  •:.i-...r 

0 OO'.;-'.  ' : 

■'  ■ r 7 i j:  . 

12 


the  distance  from  the  foot  of  the  Imee  ^racC  to  the  top  of  the 
column,  divided  hy  the  span  length.  Then  U]_  becomes  the  stress  in 
any  truss  member  due  to  the  forces  shown  in  Fig. 6. 

, A , and  E are  as  given  before. 

Values  of  U^,  P]_  i /A,  and  Pi  Ui  Z M various 

members  are  tabulated  in  Table  II. 


A 2,216.93  . 1 

dii=  inches. 

hi 

The  deformation  of  the  bottom  chord  JL  of  the  truss,  which 

is  designated  A.JL,  is  the  summation  of  the  strains  ^1^  of  the 

' AE 

bottom  chord  members.  From  Table  II, 


for  the  bottom  chord  members  = inches. 

ilraphical  He thod:  The  Willidt-Mohr*  displacement  diagram  shown 

in  Fig. 6 is  constructed  by  using  the  strains.  Pi  t , given  in  Table 

AE 

II.  The  Williot  diagram  is  first  drawn,  assuming  the  point  G to  be 
fixed  in  position  and  the  member  GJ  to  be  fixed  vertically  in 
direction.  The  Mohr  rotation  diagram  is  then  drawn  to  correct  the 
error  in  the  assuiTied  direction  of  GJ,  using  the  condition  that  the 
point  X does  not  deflect  vertically  away  from  its  initial  position 
as  a basis  for  drawing  the  diagram.  Deflections  are  measured  from 
any  point  on  the  Mohr  diagram  to  the  corresponding  point  on  the 
Williot  diagram.  Deflections  are  up  or  down  and  to  the  right  or 
left  according  is  the  point  on  the  Williot  diagram  is  above  or  below 
and  to  the  right  or  left  of  the  corresponding  point  on  the  Mohr 
diagram.  The  Mohr  diagram  is  shown  in  dotted  lines.  The  Williot 
diagram  is  shown  in  full  lines. 


*For  theory  and  method  of  construction  of  the  Williot-Mohr  diagram 
see  ’’The  Theory  of  Structures”  by  Spofford. 


a 


: t j-  'I  > 


y 


r '*1 


n 


-5  f c ‘ ■ ' : " '•  ‘ ‘ •'  h 


j -'j-.t 


< - . . . ? 


c • * 


i 


i 


j 


0 * ff  ■ I ;. . 


*.  lOV 


i J V.‘  ’ ' '. 


1 


J 


* 


n •:■ 


( 


O' 

( 


J. 


■ 


1 1 


:>'V 


;.'  ' ’.r 


r • 


« 


K-'l  ’ .W 


13 


The  algebraic  method  requires  V PiUi  Z to  bo  wiritten 



for  each  panel  point , whereas, all  the  deflections  of  the  truss 
are  obtained  at  once  by  the  graphical  method. 

The  deflections  dj  and  d(j  of  the  column  DJ  may  now 

be  found. 

dig  = 8 l/2(dii+  \ ) = inches.  (Pig.d) 

de  =4+di5  = 

dj  = dg  + dii=  — inches. 

The  total  deflection  of  B (Pig. 3)  is 

dB  = 2dG  + 2dii  + A,J1  = 8g,48S.04  inches. 

The  vertical  deflection  of  any  panel  point  of  the  triss 
due  to  a horizontal  force  of  one  pound  at  point  B,  is  obtained 
by  scaling  the  distance  in  a vertical  direction  from  the  point 
on  the  Mohr  diagram  to  the  corresponding  point  on  the  Williot 
diagram.  Then,  the  deflection  diagram  of  the  structure  (Pig. 3) 
can  be  drawn  to  scale.  As  previously  stated,  the  horizontal 
reaction  at  B due  to  a vertical  load  of  one  pound  at  any  panel 
point  is  equal  to  the  vertical  deflection  of  the  panel  point 
loaded,  divided  by  the  horizontal  deflection  of  B,  both  de- 
flections being  due  to  a horizontal  force  of  one  pound  acting 
at  B . 

hand 

The  horizontal  reactions  on  the  right,  column  due  to 
vertical  loads  at  various  points  of  the  structure  follow; 

1 lb.  vertical  at  3;  Hh  =||  =Qf^|||  = 0.02863  lbs. 

1 lb.  vertical  at  6:  Hr  = 83^f§3~"  0*03162  lbs. 

1 lb.  vertical  at  4:  HTy=^4  « = 0.03019  lbs. 

^ dg  83,483 


nps 


14. 


The  horizontal  reactions  on  the  left-hand  column 
are  equal  to  the  corresponding  horizontal  reactions  on  the 
right-hand  column.  Table  IV  gives  both  horizontal  and  vertical 
reactions  due  to  a load  of  one  pound  at  various  panel  points  of 
the  structure. 


m " 1- 

t‘'  ■'.  ' ' . ■ ;•  '’  , ff  .|||L  , ‘ 

^ n'*gr  .Cqo  ,J^{\ X ' cfo  ?iirs;.  t Jt ' xJ^  v'6%  iToxC  £»^vj  . - . 

■'  • '^  ' ' *\‘'  ■'  ' " '?}W  ’■'B!^''#'' 

I ' ■ 'K'--  • 'tf'-  ' ■ »■  .'  *'  ■ ‘ ^ 

i>£i di'^  P'Xtfxty  . d3fcwi»fe*/>^  ircc'it  • I 

^ ■■  1 .V  . 


r;.'|;e  ,'iwM>4iOii;  X^'^f' -t 


'f  ■'■• 


Vl  . r ^ 


"i 


..•■If 


■id  * ' ' "'  ^ '^'  '*  • ^^^' 


!>  ' 


r^-=* 


r •»  • 


• . L'  ^ ^ ..' 

^ ■ ' 

t 'r  ):  ■:  m:iS 

i , -I  (*.  ..v.!  jl/iR 


. ■ '<<  " • .'''/A'’ 


4't  'v"(^‘:  ■.?% 

< ;,>■  >4::-:-'=  : ' ■■  ■•'i- 


,-*f  J ‘ ' . '*  %/: 

‘ i.  ' t ' "■';. 

* T , ' 1 ' : . i , if  ;'  ' ' '■ 

F.  I . fUMi:  . ' ■ V '1  Jl  •'  < 


■V  ’ I • ‘ ' 

. t * ’'f''  ' -7'  - 

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1 > . * 1 *S'  . 

,’  .’  c " ■ ■■.  ... 

•f.'J.  ■ ■ N,,.:Mi'' 

wl, 


it  ■ ft  ' ' \ V.  .'  . 


I-  ‘ 


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■‘  '".'lA' 


. . , I '..X^'^  A i 

■ty  “l1S«)gP’*^t!l?>PSL.'U'  '.<  ■ -g.n>  j.' *W 


' ■«  V.  ’ ... 


6.  Application  to  Class  II,  Colimins  Fixed  at  Their  Bases. 
This  type  of  structure  has  six  unknowns,  viz.,  the  horizon- 


tal reaction,  the  vertical  reaction,  and  the  resisting  moment  at 
the  base  of  each  column.  There  must  be  a solution  for  each  of  the 
three  unknown  reactions  on  one  column.  Then  the  three  unknown  re- 
actions on  the  other  column  follow  from  the  equations  of  static 
equilibrium. 

Case  A.  Horizontal  Reactions. 

The  influence  line  for  the  horizontal  reaction  at  B,  which 
is  the  deflection  diagram  of  the  structure  v/hen  it  is  subject  to  a 
horizontal  force  of  one  pound  at  B,  will  now  be  obtained.  The  meth- 
od is  as  follows: 

Apply  a horizontal  force  of  one  pound  at  B (Fig. 9).  An  equal 
and  opposite  force  must  be  applied  at  D to  produce  equilibrium. 

Point  D is  considered  to  be  fixed  and  point  B is  considered  to  be 
free  to  move  to  Bg,  due  to  the  bending  of  the  columns  and  the  de- 
formation of  the  truss  under  the  applied  forces.  Each  column,  be- 
ing fixed  at  its  base,  must  remain  vertical  at  its  base.  These  two 
horizontal  forces  cause  bending  in  the  columns  and  develop  forces 
at  the  points  where  the  columns  are  connected  to  the  truss.  Due  to 
the  unknown  bending  moments  at  the  bases  of  the  columns,  these 
forces  are  statically  indeterminate.  They  will  be  denoted  by  sym- 
bols for  the  present,  as  shown  in  Fig. 7,  and  their  numerical  values 
will  later  be  found  from  the  condition  that,  at  certain  points  of 
the  structure,  the  flexure  of  the  left-hand  column  equals  the  de- 
formation of  the  truss.  These  forces  acting  on  the  columns  are 
equal  and  opposite,  when  referred  to  the  truss.  F is  the  unknown 
force  acting  at  J. 


^ a 


'•'^  ^F‘T 

B n ‘ ■ ~v,  xi^  ‘?tj  ■ V4  a/,i't  ■ 

!•  ^tjcitcuxvx  X/fwiiTl*;..  6.i#'^,»ai'/^ 

■ — " ■■  1*  .' .t . . . w:  ■ "■ 


'**  ■**'  M^'rt  rfati,  -io  tmjsMp:  ' 

r'  L j.  ~ j ! V »*A  ^>-...y.  t ‘'-^5''  ' ^■ 


/ >» 


' 7"<ki 


't  , . l''i 
' ■>  ^ 


if  - 


/ 


if., . 


^ > a I 1^  i J l|H|  r >£j  I ^ . 

^ '•*■  " ■f.'’  t'.y-'oq  eaj^i. 

, r\’  ' .\.  . ^._  ' ■.  ..  ■ >«^  ',V'^.'.^'  ,T,  J 4,ro£ifol  •> 

i-Br  .(■■’.^’5;--  / Wttir  r’f;  . 1-v  ^onsav  ' * 

,^v.:  . '..  . .:  a'.'-.  ..  ' 


f:t.i  i#  Xfcp.bt  .n*  tiilft:  « 1^‘x  iV'^i 

, 5*0-r4i  -^i'6.T^  -1*  S-ti-ia£'  ii^i0.fta)wt!i,ixiii^ 

'’’y'*  : ^ ■ fi  j-  ■ MW ' 


■■■^•'^^a.i.-H^Ui  V.:- i^fcJr-o-qo  a.^ton. 

,6a  . ..  !-.  os?*#  <i,;j  .j,,  ■ 


Jg^'igJtaLariii 


16 


Fig. 9 is  a greatly  exaggerated  deflection  diagram  of  the 
structure  when  subject  to  the  horizontal  force  of  one  pound  at  B. 

The  solid  outline  is  the  original  position  of  the  bent.  The  broken 
outline  shows  the  structure  with  its  original  form,  but  displaced 
horizontally  through  the  distance  DD^  = BE^,  due  to  the  bending  of 
the  column  DJ.  The  final  position  of  the  structure,  which  is  due 
to  the  deformation  of  the  truss,  is  shown  in  dotted  lines.  The 
point  B is  displaced  from  B^  to  Bg  by  the  deformation  of  the  truss 
and  the  flexure  of  the  column  BL.  The  deflection  diagram  cannot  be 
drawn  to  scale  until  the  flexure  of  the  columns  and  the  deformation 
of  the  truss  have  been  determined. 

Flexure  of  the  Columns. 

The  forces  acting  on  the  two  columns  (Fig. 7)  are  equal,  but 
opposite;  therefore,  the  flexure  of  the  tw^o  columns  will  be  the  sam< 
in  magnitude,  but  opposite  in  direction.  Each  column  acts  as  a 
beam  under  the  forces  transmitted  to  it  by  the  truss  at  the  points 
where  it  is  connected  to  the  truss.  The  elastic  curve  of  the  de- 
flected column  DGJ  is  shown  as  DG^Jg  (Fig. 8).  Draw  a tangent  to  th( 
curve  at  D.  The  deflections  of  the  points  G and  J away  from  the 
tangent  to  the  elastic  curve  at  D are  easily  found  by  the  area-momei 
method.  The  numerical  values  of  these  deflections  cannot  be  found 
until  the  numerical  value  of  F has  been  found.  The  moment  diagram 
is  d’g^jd. 


-dj  = 


1 


El 


[72Fx36x48  f 72Fx90xl32  + (72F-180)  x 90  x 192] 


= |j(2,223,936F  - 3,110,400) 

: 27,388. 37F  - 38,305.42  ir.ches. 

E 

-do  = |i  [72Fx90x60  f (72F-180)  x 90  x 120] 


17 

» |j(l,116,400F  - 1,944,000) 

» 14.364.53F  - 23,940.89  . , 

— I j inches. 


Deformation  of  the  Truss. 

It  is  now  desired  to  determine  the  horizontal  deflection  of 
point  J,  due  to  the  deformation  of  the  truss,  and  the  deformation 
of  the  bottom  chord  under  the  forces  shov/n  in  Fig.  7.  The  deflec- 
tion and  the  deformation  must  first  be  expressed  in  terms  of  F,  and 

tht. 

the  numerical  values  of  the  quantities  can  be  found  when^ numerical 
value  of  F has  been  determined.  The  algebraic  method  must  be  used 
until  the  numerical  value  of  F has  been  found.  Denote  the  horizon- 
tal displacement  of  J due  to  the  deformation  of  the  truss  as  dgj, 
and  the  deformation  of  the  bottom  chord  as  AgJL. 


AE 


P2  is  the  stress  in  any  member  due  to  the  forces  shown  in  Fig. 7. 

Pg  = SF  4*  Q for  any  member  (See  Table  2),  where  F is  the  unkno\ra 
force  acting  at  J,  S is  the  numerical  coefficient  of  F in  the  ex- 
pression for  the  stress  in  any  member,  and  Q is  the  numerical  part 
of  the  expression  for  the  stress  in  any  member.  ^ is  the  stress 
in  any  member  due  to  a unit  horizontal  load  applied  at  point  J,  or 
due  to  the  forces  shown  in  Fig. 5.  The  method  of  supporting  the 

truss  in  Fig. 5 is  explained  under  Class  I,  Case  A.  The  symbols  2, 

P pUi  I 

A,  and  E are  as  designated  before.  The  term  is  found  for 

A 

each  member,  and  the  sum  of  these  quantities,  when  divided  by  E, 

gives  the  value  of  isi. 

^2*^^  " ^ ^or  the  bottom  chord  members. 

The  values  of  P2,J/t  , and  ^2^1^  for  each  truss  member  in 

A A 


18 


terms  of  /:  are  given  in  Table  II. 

A1 80^  from  Table  II, 

d21  - 6p5.,5lF.f  70.5^ 

ill 

AsJL  = 200.94F^F  427.56 

Referring  to  Fig. 9,  the  following  relation  is  seen: 
dj  - dQ  + d2i  or  dj  - d^  = dgi. 

Substituting  in  the  equation  the  values  of  dj,  dQ,  and  d2i 
found  above,  an  equatidn  in  F is  obtained. 

-13,033. 84F  + 14,364.53  = 605. 51F  f 705.99 
F = +1.0031 

This  value  of  F^  is  now  substituted  in  the  expressions  for  the 
flexure  of  the  columns  and  for  the  deformation  of  the  truss,  and 
the  numerical  values  of  these  quantities  are  obtained. 


dr  = 10.858.38  , , 

j — inches. 

^21  * inches. 

E 


dr.  = 9.545.59  . , 

br  — inches 

AzJL  = .628,72. 

E 


The  numierical  value  of  the  deformation  of  each  member,  — , 

AE 

is  found  and  tabulated  in  the  last  column  of  Table  II.  From  these 
values  the  ?/illiot-Mohr  diagram  is  drawn  (Fig. 10).  The  Williot 
diagram  is  first  drawn,  assuming  the  point  G to  be  fixed  in  posi- 
tion and  the  memiber  GJ  to  be  fixed  vertically  in  direction.  The 
Mohr  diagram  is  then  drawn  to  correct  the  error  in  the  assumed  di- 
rection of  GJ,  using  the  condition  that  the  point  K of  the  truss 
does  not  deflect  vertically  away  from  its  initial  position  as  a 

basis  for  drawing  the  diagram.  Deflections  are  measured  from  any 

to  the  corresponding  point  on  the  Williot  diagram, 
point  on  the  Mohr  diagram^^^  DeTlections  are  up  or  down  and  to  the 


il 


i 


Ti  r 


,'I  c. i: : 


•■>  ^ ■’ 


:-c  • 


• U '■'  ' 

•I  '‘-*  V u».'  ^ ? 

- ' t J 

LI 

■ ..  > 

. i - : 


..  . i . * *1 


- -b  . 

- aLgC^ 


:.i-  ••  as-M 


rviV' 

1'T!’ 


y. 


;.+  fV.:  -tia'i/Qv 

■^i  .•■•.:o.":i-  " 


i'  J : : 
*■•■  -» '^rr 


' *:a 


a’-^x. 


1 ' I 


lc^il  V C • i CJ/w 


. . ....  . L?  - 

■ .i  ■z 

- * 

'»  liy  - - -v-i-*, 

I A « -^  • f* 


‘ > ’l  "i  .1 


C.f':,  -•) 


Z^.  ‘.''lu  "^O  il/ti  . 

i:'..  r --■■  ,;v  . 

, . ^ i L I !Oa,>  , '• 

rti  : ■ :.■  * s.r  c . LC-  •■:>. 

■ "u  ••  ;•  •■•..'? \^„oo  rr  /i^>;-! 

' I 

j.''"  ' * .:o  :*  r.L.  i n/i  t 

» I -V 

i-ilx  iiv'X  4*>'01*J5**  '*  L%  ~ j' . 

I-  • & 


'll. '3 

> V ^ 

■ : 

iloU^ 

.1. 


a , 


■-J  .T '.'I ‘•^>57 


• v^  * • ■ . 


e-J-5  .j  r 


s ( 


t > ; J ;;i.  *j  . ;•  " '?  .:•  .C 


: • ;,o 

• • ».  ’"  'u;  * , 

<v  I:  lj  iajBtf  i, 

Z..'A'.  '.■'.3  ,TO!' 


'■a*iantq(||l  ■ - teKjg  as*  jagtjaf 


19 


right  or  left  according  as  the  point  on  the  Williot  diagram  is 
above  or  below  and  to  the  right  or  left  of  the  corresponding  point 
on  the  Mohr  diagram.  As  before,  the  Mohr  diagram  is  drawn  in  dot- 
ted lines  and  the  Williot  diagram  is  drawn  in  full  lines. 

The  total  deflection  of  B,  Fig. 9,  is 

^ ^‘^Sl  inches 

iL 


The  vertical  deflection  of  any  panel  point  of  the  truss,  due 
to  the  horizontal  force  of  one  pound  acting  at  B,  is  found  by  scal- 
ing from  the  proper  point  on  the  Mohr  diagram  in  a vertical  direc- 
tion to  the  corresponding  point  on  the  Williot  diagram  (Fig.lO). 

The  deflection  diagram  of  the  structure  (Fig. 9)  is  then  drawn  to 
scale.  Then,  the  horizontal  reaction  at  point  B,  due  to  a vertical 
load  of  one  pound  at  any  panel  point,  is  equal  to  the  vertical  de- 
flection of  the  panel  point  loaded,  divided  by  the  horizontal  dis- 
placement of  B. 

The  horizontal  reactions  at  B due  to  a unit  vertical  load  at 
the  various  panel  points  are  as  follows: 

1 lb.  vertical  at  3:  = 0.06265  lbs. 

^ dg  22,346 


1 lb.  vertical  at  6;  Hg  = ^ = 32" 546 

O-B  > 

1 lb.  vertical  at  4:  ^ 


0.07026  lbs. 
0.06632  lbs. 


The  horizontal  reactions  on  the  left-hand  column  are  numeri- 
cally equal  to  the  corresponding  horizontal  reactions  on  the  right- 
hand  column.  The  horizontal  reactions  on  both  columns,  due  to  a 
vertical  load  of  one  pound  at  various  panel  points,  are  tabulated 
in  Table  IV,  Class  II. 


20 


Case  B,  Vertical  Reactions. 

It  is  now  desired  to  determine  the  vertical  reaction  on  the 
right-hand  column,  due  to  a vertical  load  of  one  pound  placed  at 
various  panel  points  of  the  structure.  As  it  was  stated  before, 
the  influence  line  for  this  vertical  reaction  is  the  deflection 
diagram  of  the  structure  when  it  is  subject  to  a vertical  force  of 
one  pound  at  B. 

An  upward  vertical  force  of  one  pound  must  be  applied  at  D 
(Fig. 13)  to  prevent  translation,  and  moments  must  be  applied  at  the 
bases  of  the  columns  to  keep  the  columns  vertical  at  their  bases. 
These  two  vertical  forces  and  moments  cause  bending  in  the  columns 
and  develop  forces  at  the  points  where  the  truss  is  connected  to 
the  columns.  Fig. 11.  These  forces  are  statically  indeterminate  be- 
cause of  the  unknown  moments  at  the  bases  of  the  columns.  For  the 
present^  these  forces  7/ill  be  denoted  by  symbols.  F is  the  unknown 
force  acting  at  J.  C is  the  unknown  force  at  D or  B,  acting  hori- 
zontally. These  forces  are  not  the  same  for  the  two  columns  (See 
Fig. 11).  The  value  of  the  numerical  part  of  the  force  at  each  poinl 
where  the  truss  is  connected  to  the  right-hand  column,  i.e.,  at  K 
and  L,  is  such  that  the  moment  of  the  couple  formed  by  these  two 
forces  is  equal  to  the  moment  of  the  couple  formed  by  the  vertical 

force  of  one  pound  at  the  base  of  each  column.  This  numerical  force 
1x40 

is  — g — » 6 2/3  pounds.  The  numerical  value  of  the  symbols  will  be 
found  later  from  the  condition  that,  at  certain  points  of  the  struc- 
ture, the  flexure  of  the  columns  is  equal  to  the  deformation  of  the 
truss.  The  forces  acting  on  the  columns  at  the  points  where  they 
are  connected  to  the  truss,  when  referred  to  the  truss,  are  equal 
and  opposite. 


21 

Fig. 13  is  a greatly  exaggerated  deflection  diagram  of  the 
structure  under  the  two  unit  vertical  forces  and  the  moments  at  the 
bases  of  the  columns.  Point  D is  considered  to  be  fixed  in  posi- 

I 

tion,  and  point  B is  considered  to  be  free  to  move  vertically  to 
Bs,  due  to  the  flexure  of  the  left-hand  column  and  the  deformation 
of  the  truss  under  the  unit  vertical  forces  and  the  moments  at  the 
bases  of  the  columns.  The  moments  at  the  bases  of  the  columns  caus(! 
the  columns  to  remain  vertical  at  their  bases.  The  solid  outline 
shows  the  initial  position  of  the  structure.  The  broken  outline 
shows  the  structure  with  its  original  form,  but  displaced  horizon- 
tally through  the  distance  DD^  = BBq_  due  to  the  flexure  of  the  col- 
umn DJ,  The  final  position  of  the  structure,  which  is  due  to  the 
deformation  of  the  truss  under  the  forces  transmitted  to  it  by  the 
columns,  is  shown  in  dotted  lines.  B has  moved  to  its  final  posi- 
tion B£  due  to  the  deformation  of  the  truss. 

j This  deflection  diagram  cannot  be  drawn  to  scale  until  the 
flexure  of  the  columns  and  the  deformation  of  the  truss  have  been 
determined. 

Flexure  of  the  Columns. 

Since  the  forces  acting  on  the  two  columns  are  not  the  same, 
the  flexure  of  the  columns  will  be  different. 

Column  DJ . 

The  horizontal  forces  acting  on  this  column  are  the  same  as 
in  the  preceding  case  (Fig. 8),  except  that,  what  was  the  numerical 
force  before,  now  becomes  the  force  C.  Therefore,  what  was  a nu- 
merical value  in  the  expression  for  the  flexure  of  the  column  be- 
fore, is  now  the  coefficient  of  C. 

-H,  = 27,388.37F  - 38,305.42C.  , 

J — 5T ' inches. 


.00' 


t/  .■ 


■-  .<  j 


r, 


' ■/.  '>••'■ 

VJ'^  L ■ - 

( • ' 


'!i 


•;  - 


■ ' " ■ ' • ..  .lO 

i'*’  ^ ...  ■ ..^i 

■ . ' 1 -t ., .' 

"'y'  "■ 

• /•-'  ■■■  ' . V 1/,:^^  . ••  V; 

’ •'  c/r;»,.'  : 


.»  ->y«  , ^ 

\il  ■" -k"- 


T ••  r 


' '.'  I 


r*' 


/l  *, 


j-V'-  .'  . :? 

, ^ 


1 


- ej 


j i 


-;.  ^,,1 


", ' ■ ! : : j^: 


V -*  ' 

« £ i . ^ivL  ^ 


w V<  I 


-3^ 


. ^'VJi 


.'  •)•'  . 


t/ 


••  t;.:. 


•• '••f 

...  I . ^ ^ 

i-:.  ,.  t i-. 

*'• 

,.*V 


■ * ' ,n 


I 


' -'•VJ ..’  ■/ 


^l< 

1 


X ••r  ■ . 

i’C! 


^ a. 


► , f». . . . , 

f ,*f 


•:/  9^  *-  /-  ■ 


■') . '.  1 


.^3  : „ 


■trrs^aiSi 


' ! \ 

f ' 


■t  ■' 

.■  a 


. nX 

, ’ , ■ ' ’..I1. 

‘ ■ ' ■■  ■.  ‘ • y,' A 


/ 


f;'”t.C>V 


22 


-dp  * 14,564.55F  - 25.940.89C 

E 


inches. 


Column  BL: 

The  elastic  curve  of  column  BL  is  shown  as  Fig. 12. 

The  line  BgL^  is  tangent  to  the  elastic  curve  at  Bg.  The  moment 
diagram  is  b‘k?b.  The  displacement  of  K and  ^©spsctive 

ly^  away  from  the  tangent  at  Bg  are  found  by  the  area-moment  method. 


of  point  K relative  to  point  G,  d.23,  due  to  the  deformation  of  the 
truss  under  the  forces  shown  in  Fig. 11,  and  the  deformation  of  the 
bottom  chord, under  the  same  forces.  The  deflection  and  the 
deformation  must  be  found  by  the  algebraic  method,  because  the  nu- 
merical values  of  the  forces  in  Fig. 11  have  not  yet  been  found. 


Pg  is  the  stress  in  any  member  due  to  the  forces  shown  in  Fig. 11, 

Pg  * SF  f QC  f T for  any  member  (See  Table  III),  where  F is  the  un- 
known force  acting  at  Jj  ^ is  the  numerical  coefficient  of  F,  and 
is  the  same  as  in  Class  II,  Case  A)  £ is  the  unknown  force  acting 
at  the  base  of  either  column,  i.e.,  at  D or  B^  £ is  the  numerical 


• 3^[(72F-)-480)  x (36x48+180x162)  - 180Cx90xl92j[ 
£I 

= (2,223,936F  - 3,110,400C  + 14,826,240) 


= 27.388.37F  - 38.305.42C  + 182,589.16 

E 

d^  = |j72Ff480)  X 180  x 90  - 180C 

= (1,166,400F  - 1,9-^,0000  + 7,776,000) 

» 14.364.53F  - 23.940. 89C  f 95,765.55 

E 

Deformation  of  the  Truss. 

It  is  now  desired  to  determine  the  horizontal  displacement 


where 


it/  btiM  „ * • rj.\.  A i*‘-rrrr 

‘i  . K'i  bitl  - ^ -T  .01^  VA\t,  ^ , 


.lU'T  ■ 


•* 

!► 


(•-  0<  . • > Jv’ 

. i.  - *:.: 

: 

-.ili;  I Q-. 

j*'  :0£l4:<i 

..f*  ■ 

' * i 

'■ 

• ■*  *’■  . . i** 

.t--  ”.  ~ ‘’f 

- 

. t 9 , * . . ; 1 

. 

v/ >t  ■:  ’{ 

i 

' c 

• • • i,  * . 

• 

• 

■'•  i¥>^"ai  u^... 

M i ^ ' 4.  i ' 

■ • . M » • 

V 

. > 

• » 

■do;; 

■ j .;  .' 

. -1 

.'  . ! V ' 

, :T,.  ij>  , :iic. ' ..roi  'i 

"Oj.  . ^ 

« 

* ' t > 

' .,  I- ■<  . ■ 

. '■  ■: 

■-"i^ 

)■ 

l- 

-:'^ ' 
i-  • 

. ,r  * . 

i t-  u.  r}  - :./: 

4 a 

‘ .»'.  '.ifaiy  . df  lil 

\ • 

. ■ .I  V , /: 

. 

\ 

r \ ",  -« 

■•'  t ^ 

: ••  0 Z'lr  It  ■ . : 

A , - ■ ’ ' 

« J 

nr  u 

■pl" 

j ; 

- • 

/ 

ir. 

. k 

.-0  ;-.-L  I-;.,  - ■ 

23 

coefficient  of  C,  and  is  the  san.e  as  the  numerical  part  of  the  ex- 
pression for  (See  Table  II);  T is  the  numerical  part  of  the  ex- 
pression for  the  stress,  is  the  stress  in  any  member  due  to  a 

horizontal  force  of  one  pound  applied  at  K.  To  keep  the  truss  from 
moving  to  the  right  a horizontal  force  of  one  pound  must  be  applied 
in  the  opposite  direction  at  point  G,  for  the  relative  movement  of 
points  K and  G,  due  to  the  deformation  of  the  truss,  is  desired. 
Thus,  U3  becomes  the  stress  in  any  member  due  to  the  forces  shown 
in  Fig. 14. 

Z,  A,  and  E are  the  same  as  before. 


The  quantity 


is  found  for  each  member,  and  the  sum  of 


these  quantities  is  divided  by  E to  obtain  d33. 

^3*^^  ” ^ f’or  the  bottom  chord  members. 


•AE 


Values  of  P^,  U3, 


P3U32 


for  the  various  members  are 


tabulated  in  Table  III. 


d33  = 1.411.97F  4 1.839.55C  4,706.60 


E 


A^JL  = 200. 94F  4-  427. 36C  4 670.09  . 

E 

The  numerical  values  of  F and  C will  now  be  determined. 
From  Fig. 13  the  following  relations  are  seen: 


dG+  d 


33 


or 


dj^  - do  - d33 


or 


“ -43JL  = 0. 


Substituting  the  values  of  these  terms  given  before  in  the 
above  equations,  two  simultaneous  equations  in  F and  C are  obtained 
27,317.09F  - 49,721.100  -f  91,056.95  = 0 
54,575.aiF  - 77,038.200  -f  181,919.07  = 0 
F = -3.3333  and  0 = 0. 

Now  these  values  of  0 and  F are  substituted  in  the  expressioiife 
for  the  flexure  of  the  left-hand  column  and  the  numerical  values  of 


24 


these  expressions  obtained. 

dj  » 91 . 294.58  inches.  dp  » 47 . 881 , 77  inches. 

E ^ E 

These  values  of  F and  C are  also  substituted  in  the  express- 

P'tl 

ions  for  the  strains  of  the  various  members,  . , and  the  numierical 

AE 

values  of  these  strains  obtained.  They  are  tabulated  in  Table  III. 
From  these  strains  the  Williot  diagram,  Fig. 13,  is  drawn,  assuming 
G fixed  in  position  and  locating  J a distance  dj-d^  to  the  right  of 
G and  a distance  equal  to  the  elongation  of  GJ  above  G,  Here  there 
is  no  error  in  the  assximed  direction  of  GJ,  and  it  is  not  necessary 
to  draw  a Mohr  rotation  diagram.  The  deflection  of  any  point  is 
measured  from  point  G,  and  is  up  or  down  and  to  the  right  or  left 
according  as  the  point  is  above  or  below  and  to  the  right  or  left  of 


G . 

The  vertical  displacement  of  the  truss  at  K relative  to 
point  G,  due  to  the  deformation  of  the  truss,  d35  (Fig. 13),  is  given 
by  the  formula. 


^35  “ 

The  values  of 


I 


AE 


for  the  various  members  are  given  in  Table 


III;  they  are  the  strains  in  the  truss  members  due  to  the  forces 
shown  in  Fig. 11.  Ug  is  the  stress  in  any  member  due  to  a unit  ver- 
tical load  applied  at  K.  To  prevent  vertical  movement  of  the  whole 
truss^  an  upward  vertical  force  of  one  pound  must  be  applied  at  G 
(Fig. 15).  To  prevent  rotation  about  G^  horizontal  forces  must  be 
applied  at  G and  J to  form  a couple  whose  moment  is  equal  to  the  mo- 
ment of  the  couple  formed  by  the  unit  vertical  forces.  The  numerical 
value  of  each  horizontal  force  is  equal  to  * 6 2/3  pounds.  ^ 

then  becomes  the  stress  in  any  member  due  to  the  forces  shown  in 


■At 


Tit 


J: 


>TKIMg^i.s.>';.iiiic«i r;ttBli:.Xir>”iii^-;^^  .gn-,  ?js:r 
4''0»  , , I ■;  . •»  , ■■■  « »•-**«'  tm  . 


5 :.  ..i\'i«''^ 


f:r 


'tr 


VV. 

Jp-  **^  ' "'“JT  ’* 

-■  A7 

:7.t>  criij  'f  i.;.Mi^iif iifffeiii ■ ^jhU^  5»«iH  'ivigi- i63aii?.ir^4d'4t^^^ 


fr  ’V I 


^{tveiW  t-if*  ^>i>4^§i.‘'4i#i;ii  i,  -o^‘57tv 

• ’ -V  , * ' p 

^ l 9l^'  X : r,  Xv^r  ^’%  u::;rttt^f^ 

• . ' • ' ■ if  . ■ ' >■  ■ 


f,  ' 'i.'.  -\*  v^l,  'W 

;5^  . A?  Kit^yu  oi>;iJt;h^,ii  5 ;5i  s.7i:at^c^X  l>ae 

C ‘ i '’? '.  -;5  - ,j..  V ’ ' , j .,  . V , .^''■^  iijf* 


» • . 3CMk^  ‘ V '!''•’  '*\i 

|..  ' . ‘.'  '*:i2  '■'■  .6^ 


r ' vXiTrJdXWij  - b's  ' 

Ck-‘-'  ' ' ’ -Xi  . ^ , ■ '"^  :m'-  '■■ 


W,  lyti' Tiv  ^pi^  .of  iifc/  poit.^  •.;■’■  pprtSp'^oi. 

*T  'k  ' * / VlBH  . . r rtl.*,'.  /^  .'. 


ni' .&^i-hk  «4X-.  QOuix^x*  .’^o.^esfi/Xxsv,  -tilt,, 

f£-'  .'  , t ',  ^ -r'-^'-  ..A  , , ' 2 ■ - ■ V^'H- 

■S'-  • ■ . ' '.i  ■■'  •(»''  4-  ■ # .,  ' . • . ' /"•* 

4.  >!>t{;;l.  £ {;*  Ci*  Ci 7 •><ir.; 8 5:  •*' i ui4??, ertt  atij  . J £I,i 

L-V:V  xrui.  nt  .CSp  v-r^tj  ni*'^  %i:;  ^!3  . XX.^tAi  I 

t -.  .Tfc__ '■  ' ' ‘ • ^ 

i,) 


OJr  Xaijjp:*  -^i,.  fcg^Iq^^ 

'■  i . }.•'■ ..®  '■-' ■ '■r^.\  ■ '■'  ‘■'*'  ^'  ■ /:  ’ .'  . *i.i^  % 

eolxfrflu/it  tjri^  9iiJ[  \(^‘  a-naijf^ 


:«V'fr 


V 

v.tr 


Jc^'-- 


r**  * 


^ a^‘X^3.0p<^ ‘.d£j''&;OioA  X4^jw:>i*x6'4l- 6Vje.®i  IsO' 

■ ■'  - ^ A'- iv, '/  V . 

rpi2<v'''  .■  '.»'  ''  .-  .■'■  ■;=*)?*’  yg  .'  i M' ’•*'  '1  *;  *,  ..  _,., ’'*'  , y^*'  V.^'.t 


.^•v; 


lu  •■  ■ •■ 

jyLaarj'^iirttirw^  '-V  ^u.  w®i^'-JJt»i ; 


25 


Fig.  16.  Values  of  Uc  and  for  the  various  members  are  tabu- 

A 

lated  in  Table  III. 

^33  = 7,48^ 

The  term  dj-d^  represents  the  value  of  the  movement  of  point 

J of  the  truss  due  to  rotation  of  the  truss  about  G.  See  Fig. 13. 

^J~^G 

The  angular  rotation  is  practically  equal  to  ^ > “the  value  of  a 

small  angle  in  radians  being  approximately  equal  to  the  tangent  of 

the  angle.  The  angular  rotation  of  the  whole  truss  about  G due  to 

dj~dG 

the  applied  forces  is  then  — ^ ~ . The  vertical  deflection  of  point 
K,  or  point  B,  due  to  the  rotation  of  the  truss  about  G^is  then 
equal  to  ^(dj-d^) , the  chord  subtended  by  a small  angle  being  ap- 
proximately equal  to  the  subtended  arc.  The  vertical  deflection  of 
point  B due  to  the  deformation  of  the  truss  at  K is  d3g.  The  total 
vertical  deflection  of  B is  then 

dg  » 6 2/3(dj-d^)  f dg5  - 296,899.47 

The  deflection  diagram  of  the  structure  (Fig. 13),  due  to  the 
vertical  force  of  one  pound  at  B,  may  now  be  drawn  to  scale.  The 
vertical  deflection  of  any  panel  point  of  the  truss  is  found  by 
scaling  from  the  Williot  diagram.  Then  the  vertical  reaction  at  B 
due  to  a vertical  load  of  one  pound  at  any  panel  point  is  equal  to 
the  vertical  deflection  of  the  panel  point  loaded,  divided  by  the 
vertical  deflection  of  B,  both  deflections  being  due  to  a vertical 
force  of  one  pound  at  B. 

The  vertical  reactions  for  unit  loads  at  various  panel  points 
are  as  follows: 


1 lb.  vertical  at  1:  Vj^  = ^ “ 0.1532  lbs. 


dg  296,899.47 


’SS  J V , , . 

:=j5j^i.aaca  jiL>ti« 


9 r S^Vi' 


^■V'v 


ii " y^'ii . 

" -'iS^  ^ 


C;/tri^lfV  ^ tJ^ isuyCa"/  .2  C.>t4' 


i.  V: 


“*  ■ *v 


= as*??*-  ■'^.r^lli-1  ^ 

■'■■»•  ^ -jg  - i ^ 

1 •, i ' 1 -WJ  . - -.-vVtr 


V - ' •■  ■'  ''■'’•r'.V 


' ’"^Vf ' 


P ‘‘i^W  fiEfc  trlodh  to  dL'Dx'd^fj'x  ,0iPZ 

I \ ^ ' ■ ' /'  ^ ‘ ■‘•"'>  ■■  * 


.■ 


/..' ' <el  ■’.d'jjicri-':  ts</W<,.  '^■a  -r^l'-:/a#o,':' 


liod  .•■  ' Vi  1 TtO.*!??.  pet-,  a 


. •;  ii>  #<;(«»■  9if^ ‘ . Ij't’ ' ^ 


T ‘ 


vitf  , ' ,aid^  ..■!  X h.  tt^.vti  (X  -tisio^ 

t : ^'  •’  ■ ' '•■V  :.  i" 


ih 


'.  .v...": ..  ■ ^::i' ■_-'#» 


3'^»'3 

«2I 


».,  '3 


. ■ mm  :, 

^:v  .:’  ■ ■ :V/  _ ..  ' ^ '..  ',  • 

^ »d'T  .ftMCta  a4  /rariu  Ji'  l **'-'«  \d'  ^ ‘t*  JPt  J^ip^f^v 

Sr .»  .'  ...i’  it*^  '•'’^  ■ *■•“.•  , ' /V.  ./•  »tV  > ,*^  ja'is'- 

..**  '■  *.  ■•  ^ ■ ■ , , " ■*  ‘ ” ■ III-"  '^.  . ^.'.:  ■ ...  ' ' >4 

*: ^ - *^‘  kniv  I lAd t 'All*' 

^ '•.'  f?"  v:  ' . „'.  !6  ■ ■■  ■;■'''*  - '■  '-^jt 

' * IV  ■ ' • ' ^ ^ */  -••  **  ^ ^V|  i],"’' 

!jltft,' tphi^df.  fdlm 

' . ;’ a i d 0 i f'tfj^  *PW  \t- tdgjr^l f ^ 

• ^ ^ ^ *v;  '.4  ,'v  ‘ , 

' . . 1 . V - ...  ■ ■,  f-  i'f.  -.--  — — r '.  ■ 

. iau.  i^'f:ii;4  i#  a 'w^ ‘ 

K*  ''.li:  i -SSSifib,  “■.:(, ''^.Si 


0:t 


V -'F 


#■  ■ r/V.‘ 

:4wol^o>  a^.,%^ 


' ,’i;  ^ ■ 


26 


1 lb.  vertical  at  4:  = 0.6897  lbs. 

The  vertical  reactions  on  the  left-hand  column  are  found  by 
equating  all  the  vertical  forces  equal  to  zero.  The  vertical  re- 
actions on  both  columns  due  to  loads  at  various  panel  points  of  the 
structure  are  tabulated  in  Table  IV,  Class  II. 

Case  C,  Resisting  Moments  at  the  Bases  of  the  Columns. 

The  influence  diagram  for  the  resisting  moment  at  the  base 
of  the  right-hand  column  will  now  be  determined.  As  it  was  stated 
before,  this  diagram  is  the  deflection  diagram  of  the  structure  when 
subject  to  a moirient  of  one  inch  pound  at  the  base  of  the  right-hand 
column. 

Apply  a force  of  ^ pounds  at  the  end  of  a rigid  lever  of 
length  k inches,  rigidly  attached  to  the  column  at  B (Fig. 19).  A 
moment  of  one  inch  pound  at  B is  produced.  This  moment  must  be  re- 
sisted by  a horizontal  force  and  a vertical  force  at  the  base  of  eacl. 
column,  and  a resisting  moment  at  the  base  D of  the  left-hand  column 
to  produce  equilibrium.  These  forces  and  moments  cause  bending  in 
the  columns  and  develop  forces  at  the  points  where  the  columns  are 
connected  to  the  truss.  These  forces  (Fig. 17)  are  statically  in- 
determinate, because  of  the  unknown  forces  at  the  bases  of  the  col- 
umns. For  the  present,  these  forces  will  be  denoted  by  symbols,  and 
their  numerical  values  will  later  be  found  from  the  condition  that, 
at  certain  points  of  the  structure,  the  deformation  of  the  truss 
equals  the  flexure  of  the  columns.  C is  the  unknown  horizontal 
force  at  the  base  of  each  column  and  is  the  unknown  vertical  force 
at  the  base  of  each  column.  X unknown  force  developed  at  the 

top  of  each  column.  The  forces  acting  on  the  two  columns  are  not 


I 


'"v  V- 


S^.-U\ndr*tti'-  ^4  ^ 5«>:  0n^i^n% 

*«f  *"  ’’  ■■"'  '.  'tt  ‘ r / .i  ;i^'  /^f  ^.wtv 


■^  '""  * ' ’ " , ’ ' ' ' „ l ' *' ' . 'W'J^Xv  ‘ ' 

iU -'i^t^y,,  , *u:i  X j5*i  ,'j!i*0'2c^ 


n <|ft  •• 

r 


fi^  44*i^t'  «ii*i  ^so 

■ ' :■■.■-  ■ . ■ ' ■ *:  »rv  .> 


' fi' , 

r£  ' .-''•'W 


t*  • ^ : • ■ 

>•  • ■ ^ w«lk-; 


' , , . •i*  'ii»t  .ijw.  ■*  • i-  i 

Xfi-r  ..' ; i v.»  ♦<<  flo/s  Xiir  ;jr^fl^^oa^ 

“'r.  .*.v''  ■ .'  : SM'^  ./-a  ':  ■ ■ ^ 

■swy.  iv ':_ . x‘  tj .-■ . v-4 % V ~ ‘''P i^JJ:  v i.ca.i^  » 

> <■•  >\*t  •••  .'-  ' > ’;  * ' ■ ■ •i’*'\l‘?!fi  *’  '''*'••  ; .jj 

'"'Si'  •ir-jn.i'i-z  <5:U  ■;«'  t>J,i  «irto  lb  ci^  yf^ttCpa:' 

X '■'  ' ^ ir^  " ‘'  ■-  \il  *iSS3f^ 

’ ' ■ ' 'fXiM  . V ;:"'  ■- 


'’!*  4 \ tl  in  t vXjb ijtt 


e'‘4 


,-i i:  w:  ti. d*i»  4041  ^ii?r  -‘i  i 6al?oq^/(t>4t^«^3A^JAP& 

o/'-^  -o  5s.-  Of  D ^ .6t--'*.iAi-’fi^’i4v.  •;(« "' 

E^V'’  ■ '.  '^'''*  " : '^'r  ;^r'  'p-  ? ' 

-'  ' -.'  . ’■  •■  . 'X.  . ■ • • ^^  h^' 

‘ ’,  * . ■.«'/■■.  ' '■  ■■  '■:^  ’ ' ' " ‘ •"'  ' ^ '".;  •J*‘.  i-  . 

p-ii*  tin^%q  W4f,;  Xi'tv,  ..t^  <:ioXt>^«?6',-  tnaiiC-OO 

' ■"■'  , C^x . . «&-V' *9^ fe'ti^Oifcfitc«rP 


-"-rrt 


K.'.. 

r-x.r'  0:rfjr-  l,y  t :T  JS  :^. '?»'-JT>:i')b 

‘ . „ ■ ■"'*/  >’  '*•  ;•  V ' *'  XN  >*,_.  “V*  >.  • ■ 


d>iOCff.V;*  'Citf'  1>4. 


\t, 


^-,v  ©ar.*’  *.io x*? ' 6?  tlJ:v 'a4>jyJi:^Y,'^:x40^ri^^ 

■'■  .'■  ."•'  ■'•’  ' • ■•  ])i  %,\'-'^  t'  ■!  ■'■k’‘'  '’r 


Jti'i 


•r^  4rjroj;Aaw,  2'^  dTl#‘iXi4^^ 

•■-'■'*  ' '’•  '.  ';■  ■»  '!■  ' X;-n  ’V  ’ 

Ipo'i-'  X iiir^.tjTjXx^"V  ;wGi),^f(X)r  iL'  Xr.uIO’i-  X«i  ^4T*  aA^44-^oiol'-i 

< '-  -'  '•  _.•  r ■ ,,  ,-  • ■'  ■ •'  -•  IS  • ^,  -'  »,  . T ('-'-'ij.^-^u  ■•  ,.■ , 5v  V 

• . ■ ,>v  ' ' '''•'  - 1 

■,jB  .nsuxXco  rfsv'^  1 


X jB 


^-'  *■,.*  y % 

d"3n  4n3^Xcp  :pf^ 

t<AK 


-I*  ' f*\ 


' ,'  ■ ' '"i  = 4 ■ ^ 


qpi 


.4*  ' * 


* i\l 

% I . ’ ■ i1 


27 

the  same  (Fig. 17).  There  must  be  two  equal  forces,  one  acting  at  K 

and  the  other  at  L,  to  form  a couple  whose  moment  is  equal  to  the 

moment  of  the  couple  formed  by  the  vertical  forces  R,  thus  preventing 

rotation  of  the  truss.  The  value  of  each  force,  as  shown  in  Fig. 17, 

is  m 0 2/3  R,  The  forces  acting  on  the  colurms  at  the  points 

the. 

where^ columns  are  connected  to  the  truss,  when  referred  to  the  trussj 
are  equal  and  opposite. 

Fig. 19  is  a greatly  exaggerated  deflection  diagram  of  the 
structure  due  to  the  moment  of  one  inch  pound  at  B and  the  resisting 
forces  and  resisting  moment.  The  final  position  of  the  structure  is 
due  both  to  the  flexure  of  the  columns  and  to  the  deformation  of  the 
truss.  The  left-hand  and  the  right-hand  columns  are  considered  to  be 
fixed  at  D and  B respectively,  and* the  left-hand  column  must  remain 
vertical  at  its  base,  as  the  conditions  of  this  case  presuppose.  The 
right-hand  column  is  free  to  turn  due  to  the  moment  of  one  inch 
pound  acting  at  B.  The  outline  in  full  lines  shows  the  original  po- 
sition of  the  structure;  the  outline  in  broken  lines  shows  the  struc- 
ture with  its  initial  form,  but  displaced  horizontally  through  the 
distance  DD^  * BB^  due  to  the  bending  of  the  left-hand  column.  The 
final  position  and  form  of  the  structure,  shown  by  dotted  lines,  is 
due  to  the  deformation  of  the  truss  and  the  flexure  of  the  right- 
hand  column.  Point  K has  moved  to  its  final  position  K2,  due  to  the 
flexure  of  the  left-hand  column  and  the  deformation  of  the  truss. 

The  point  K*  (Fig. 19)  is  the  point  where  the  tangent  to  the  deflected 
column  at  B cuts  the  horizontal  through  K,  or  it  is  the  final  posi- 
tion of  the  end  of  the  lever  where  the  force  of  ^ pounds  is  applied. 

This  deflection  diagram  cannot  be  drawn  to  scale  until  the 
flexure  of  the  columns  and  the  deformation  of  the  truss  ha.ve  been 


igr.3.k~Jas:«r.  .jAiv,  -aia 


gsato-f-aste^^ 


TV  W'  ' 

\ ' - ■■(.  ; V 

.V,  kiti3''iw5 


% 


;pv'l^r\- $/*':?>»  jj/c  ■•'  'ft  w'ft’vorfC*’  ’ | 

!l  ^ni;t  o^  ,iM  iA  •' ii?  riqaot^^A  v'’^ 

JL'  I ■ / ' ,.  ..fj--  '■''^  ■ 0 

F;rjc.  n ^^»oTo,i  i^oji.'  ■xo  ^at  ‘ 


■ > ■ --J  »•  , !■'.'■ , ?'■  .5" 


3 


r»  I .^*7  !,»"  ■ 


la^/s;^  04  b^Jxjrr.1  » \ ,^^’/yj'  ftliJ  *c^i*  ®t#  ^fiBa/Ioo.&tsfrvJ 

■""  ■ ' ■■  "^  :;  > "? 


'""^  'M  . *.  ■'■■  >>■  ■ ^ pS'3 

' ■■  ^ thy  l4^.pf p: ' ! 


• V '“V  - ■ ' -Vi;;. -.„i-  ./^•'  ^Yy 

a:it  ao/.<^C*XA.r  ■ ':Xj>:4^»as  1^  ii'.?5rX.sil  ‘ J 

r*'  ' ^ ■ ...  '■  .■■;  ■ '^,  ■'  ~Js 

{itS^iosx  oiU  ikiAM  X-'  :<0'iJ  ^no  ^o^tx^'icm 

K"'  ' . Vv  " : 

•»  i :-t  ?•  l \\iXja4^ '' ' ■'  ' i-4-^Xfli  XSifif  c'iTK^^ 

rt.  ’ ■^  ^ ■ • • • ’■•  I ^‘’v,'»\  " .p^  ‘ ’'  f •.  1 4 ''A**- 

_'i^  n^X^iaiilV/  «ir»  ..  ,-.i  -jXiU’t%  s.tj  lc»  •'fesce*^.  ^ 

■‘  ''  : . . h ■■'■■'’  ....  " 

f>fiZ,  . i«4v2f<7q;^ap«Y  ,«io  fcj^ii  To.v. : tTj®S’-i;v  ^,rix  tit v\ osis^ 

'.'■'  ■ , ■'.'  ' . •.‘;f‘r.-* 

f ■ ’ v.'oAi  s.'f.o  I'j  lrt#4ort -liieft^ 'fiiji  .ft^^  a!^a^  nmu^to.o::^'^tLMfr^s^i'x 

^-  ■ • > ' '■'‘'  * ^K'*,  ■ " '''  "V 

~:5<r  fiftifXi/  XXxV  ’‘.  dA'i  ' 

& ‘i  ' 


R'  TQp,'-  < -^ 

pi?3ti  vk^alt  .miorid  At 


I MS  .J^-i/S?  yXXftt  iWo!:  njil, 

^ ' ■■.  ■ ■/'  ' B ' '^  . ■ ‘ v-f^  • ‘:? 

•■pJfrMV  •'  •«*JtOii-tA.,.fy#^'‘ta  t.tpi  .f?«i;/50;£;fXapqr>X*iip 


.C.'Ju 


H 4|t..i>^x‘vnv  !i;;^  zt  xei,  aC'iit4iSffot&fi:;&it3'(5^ 

My  \ rii  :X<- XxjO'V  I uit  fce  7v*c#  bM  3fi';/nli^,^;,^^^y 

’■  ' ,l;'\  ^ ':  : V.'  -.  ^ '■'  > <'  ' *^‘' ; 


? Tr^-y-~ggTcgrar»MrgTyg!:^^ 


28 


deterir.ined. 


Flexure  of  the  Columns. 

Since  the  forces  acting  on  the  two  columns  are  not  the  same, 
the  flexure  of  the  two  columns  will  be  different. 

Column  DJ : 

The  forces  acting  on  this  column  are  the  same  as  in  the  pre- 
vious case  (Fig. 11),  except  that,  what  was  the  numerical  vertical 
force  before,  is  now  a vertical  force  R.  The  flexure  of  the  column, 
being  due  to  the  horizontal  forces,  is  the  same  as  in  the  previous 
case  (Fig. 8). 

-dx  = 27.386.37F  - 38,305.42C 

E 

“dft  = 14.364.53F  - 33.940. 89C 

E 

Column  BL: 

The  elastic  curve  of  the  deflected  column  is  BK2L2  (Fig. 18). 
Draw  TKgU  tangent  to  the  elastic  curve  at  Kg.  BK*,  the  deflected  po- 
sition  of  the  lever,  at  the  end  of  which  the  force  of  ^ pounds  is 
applied,  is  tangent  to  the  elastic  curve  at  B.  The  moment  diagram  i£ 
b'k2b.  The  deflections  of  various  points  of  the  column  away  from 
the  tangents  TKg  and  BK'  are  easily  found  by  the  area-moment  method. 

a Iy  f(l-180C)  x 90  X 120  + 1 X 90  X 60)J 

- 1_(16,200  - 1,944,000c) 

El 

» 199.51  - 33.940.89C 
E 

^ £(1  - 180C)  X 36  X 43] 

El 

■ (1728  - 311,040C) 

= 21.38  - 3.850.54C 
E 


r\\ 

ktJpWlCt 


.V  t ’ 
.<  X 


- 


J-O': 


V « ,. 


'f  ^’IP  «T 


-i. 


• O'J 


, ■'Vi»E' 


COVOlI  XxiDi?10< 


tl  vz ..  , I - 

. ;. : ! r 

O.J 

;.i 

,\ 

1 :J;>0*.  . . 

i 


» V i 

/ 


« *s 


kUf£^:- 


o : 


p: 

L*V' 


F 


..■■  j .;.  .IjSk-"  - » 


■V  . 


v'^'3 


<■■  .-'•  ) 

V ^ ‘ 


•-,  -» ,i  'I 


r.o‘  ' • 


f ' 


• ^ '•.  ■■  • ' ■••  7l! 


» .■  j'\.  ’>  V%  ti%fi , : H'.  ' 


■-  I 


.-ri 
A i -l 


r _ j~~  ux 

TV 


ff: 


p-  -'-  x:3  -n  ~ . t- 


\ 

/ ™ 


-.aoRess;^' 


] 


29 


^4  X 90  X 60  4 1 X 90  X 12o] 

. 2^(16,200  - 972,0000) 

El 

» 199.51  - 11,970.440 
E 

5 =:  ~ jjl  - 1800)  X (36  X 48  ^ 90  X 132)  4 / x 90  x 192)] 

m 1-  (30,888  - 2,449,4400) 

El 

a 380.39  - 30,165.520 
E 


Deformation  of  the  Truss. 

The  displacements  of  certain  points  of  the  truss  due  to  the 
forces  shown  in  Fig.  17  will  now  be  deternriined . These  displacements 
must  be  found  by  the  algebraic  method,  for  the  numerical  values  of 
F,  0,and  R have  not  yet  been  determined.  The  displacements  desired 
are  d4i,  the  horizontal  displacement  of  J relative  to  G,  due  to  the 
deformation  of  the  truss;  d^,  the  horizontal  displacement  of  K 
relative  to  G;  and  ^^JL,  the  defoririUtion  of  the  bottom  chord. 


^41 


= I 


P4U1Z/ 

AE 


A^Jh 


for  the  bottom  chord  members.  is  the  stress  in  any  truss  member 

due  to  the  forces  shown  in  Fig. 17.  = SF  'f'  QC  TR  (See  Table  III) 

F,  C,  and  R are  the  unknown  forces  explained  before.  S and  Q are 

the  same  as  the  coefficients  of  F and  C in  the  expression  for  P in 

3 

the  previous  case.  T is  the  numerical  coefficient  of  R and  is  the 
same  as  the  numerical  part  of  the  expression  for  P^  in  the  previous 
case . 


U2_  is  the  stress  in  any  member  due  to  the  forces  shown  in 

Fig. 5.  Ujls  the  stress  in  any  member  due  to  the  forces  shov/n  in 

P4U3^ 

Fig.  14.  The  termi  — is  the  same  for  each  member  as  the  term 


30 


P5^5^  in  the  previous  case,  except  that,  what  was  a numerical  value 

AE  P/iUt/ 

before,  is  now  the  coefficient  of  R.  Values  of  ..  for  all  the 


members  are  tabulated  in  Table  III. 


AE 


(i41  =*  605. 51F  + 705. 99C  t 865. 90R 

E 

d43  a 1411. 97F  4-  1839.350  f 4706. 60R 

E 

44JL  a 300. 94F  + 427.560  f 670. 09R  . 

E 

It  is  nov;  possible  to  set  up  three  equations  containing  F, 

£,  and  R,  which  may  be  solved  simultaneously  to  obtain  the  values  of 


these  unknowns. 

(See  Fig. 19) 

- 

dQ  + 

dj  - do  - 

• d4]_  * 0 

(1) 

d^  f d43  or 

dg  - dg  - 

0 

II 

to 

(3) 

The  third  equation  may  be  obtained  by  taking  the  column  BL 


(Fig. 17)  as  a free  body  and  writing  summation  moments  about  B equal 
to  zero. 


1800  - 72F  - 480R  - 1 « 0 

p - 1800  - 73F  - 1 . (3) 

480 

The  value  of  dj^  may  be  expressed  in  terms  of  known  quanti- 
ties ( See  Fig. 18)  . 

d^  + ki  4 = y(dL  t ^5) 

7dj^  - ^“^5  " *^^4* 

Substituting  this  in  (2), 

5(dj  f 4^JL  + ^5)  - 7(d(j  + d43  + A4)  » 0 (2a) 

Substituting  the  value  of  R from  equation  (3)  in  equations 
(1)  and  (2a)  and  reducing, 

+13,499.47F  - 13,333.840  - 1.80  = 0 
f40,829.88F  ^ 64,930.500  - 567.08  = 0 


31 


Solving  these  two  equations  for  F and  C and  substituting  the 
values  found  in  equation  (3)  to  obtain  the  value  of  R,  it  is  found 
F = +0.005404;  C = +0.005336;  R = -0.0008932. 

These  numerical  values  are  now  substituted  in  the  express- 
ions for  the  flexure  of  the  columns  and  the  deformation  of  the  truss 
to  obtain  the  numerical  values  of  these  expressions. 

■ ^.^^inches.  -inches. 

iii  IL 

The  values  of  the  strains  in  the  various  members  of  the  trusi  , 

^4?  , are  computed  and  tabulated  in  the  last  column  of  Table  III. 

AE 

These  values  are  used  to  construct  the  Williot-Mohr  diagram,  shown 
in  Fig. 20.  The  Williot  diagrani,  shovm  in  full  lines,  is  first  drawn 
assuming  the  point  0 to  be  fixed  in  position  and  the  member  GJ  to  be 
fixed  vertically  in  direction.  Then  the  Mohr  rotation  diagram, 
shown  in  dotted  lines,  is  drawn  to  correct  the  error  in  the  assumed 
direction  of  GJ,  using  the  condition  that  the  point  K does  not  de- 
flect vertically  away  from  its  initial  position  as  the  basis  for  con- 
structing the  diagram.  The  deflection  of  any  point  of  the  truss  in 
any  direction  is  the  distance  from  the  point  on  the  Mohr  diagram  to 
the  corresponding  point  on  the  Williot  diagram  in  the  direction  of 
the  desired  deflection.  Deflections  are  up  or  down  and  to  the  right 
or  left  according  as  the  point  on  the  Williot  diagram  is  above  or 
below  and  to  the  right  or  left  of  the  corresponding  point  on  the 
Mohr  diagram. 

The  values  of  F,  C,  and  R found  above  are  now  substituted  in 
the  expressions  for  the  following  deformations  to  obtain  the  numeri- 
cal values  of  these  deformations. 


■'r  fim ■▼  .1 

i'  ^ 

'■j, ' <t..  r^- ,’v!i  I. 


' .If 


%*»r  • » : r..  ‘»,xyif  ^lacii 


. -r  . ^ 


^'v 


-^5  r-  ■ -i^  e'VAi  Iicxt3im.rx6  ^.©AT 

^‘-  „ ■ ,..  ■ - >n  , . i,'  y ^ ^ ^ 

• .^x.ni ^ 2v  T"  ildii  ik^\^dt:  AjU^-ci'o  • q^' 

"N  A ' 

L ' ^ - ,i,  , _;'  ift  ■ -'-. 

f .1.x  ttl^zX  ^‘4yuy^-‘ s^ofcl  fe»4it 

. " . _,  ' ■-  ""  '.  /• ' "'  . ‘<iA  . 

. ..if, * ofl c , i^i. th  n ti  3k.-xt.d I ZLl A 9>i;J . j oj ao o 0 1 t,  j:  e 'p);  a 04i X/yvc^a 

i^'  > ■ , , ^ t^-  • , ■ . , ,.-■'  ' :'  '■-(^  ■ ^ 

A,i  iViif.xcr  t|ju^  i^iA txf 

t^.|aoq  ^?Jt^’'-i-;facir  /ac^ 


y.3tn  dai^-?  ;A.;l/;rt.:xn:y,_  wl<r  , 

r.X-iAr^  3,3  Vi’i’  Wtl  ■^XXAqi'f ‘lav ■'^ii 'i 


■^■df-tt  x,U'X'3  •>;x!jt‘-  ■ ai’  fcijBii?'  vi  ,$& -iif  £>e#i^.  ’ 


.•■o;  de^'V  idt  '%<^  fSilo^ 

^tiV,----  - - ■ - ^ ■■  ■’ 


?,ju>‘  s»«5S4^<X  t • z fol^  .,pdi,  c _ 1 a I 8» t An  if  • 

j^io,.  ji'>Xifp-crf  .3  ito,  if  ft  X ^ '•  . 


• ' ' I " i‘  l.-  ’1 

bM  i#'i|3  *:i'*"''.X  i’.>*  px.;  a.vfK^TC^oXji'SjCI  . 




m 


- , .V'  ] 

■ '7*' ••■;.&:^  ' ‘1 

>•  ■ *i  ‘',  J 


L a 4'  iJUjT^'^4/:;  J'x  l.Jx‘'Xr(  :t.i>  'SiS  */U.6xp,4o4‘'‘4'l 


.10  :,  ■ ri^*-'^a.  t'Ml'.'tJt>  2«4iri^T  a 

K f :.:^j  ; 

W',  . '■'''  ‘ '“f.h'  Ji  .i^Vi^ib, 


rt  1 i a ’/^^^  >1 

I.  .■  ’ 

I*" 

^f- 


'v.viiv'^  'yyik  ':^  ,^  « 
j?,  'f  aM|i^lfi!f''-b3t^’-,  ..0;  ,,'r 


r« 


xtiii  r*-tpt4:'  v-lf  it 2 ■‘.f^i''U  P'X  '.-*>*  .?.1X  ' tqt  ii.«ioial'd--!|tm 

■f’f  . ■-  * ■ ■ jf  ' ' \'-’  '^'■X 

.,M’  *'  ’ « fiiik*Mi;t:'  la  ooyXiv^-Ica'-i 


j:  . 


.1^  . A'9i 


iB,se^^ij^tLJ9MJif9Kb£x^'.w^  .?crr  :ayii'Tr-']r»^,Jtawy^ 

t ’ . _ I - 1 * ji*S  - 1 , u . i'  '1  •'  ii  - A 


iil 


i V*' 


32 


d4i 

a 6.27 
E 

inches. 

II 

71.76 

E 

inches. 

^43 

a-  13.24 
E 

• inches. 

42  - 

-0.84 

E 

inches . 

4JL 

S 2.77 
E 

inches. 

A4  - 

135.63 

E 

• inches 

The  deflection  of  the  tangent  is 

dj^  + ^ = dQ.  -f  d^2  + ^^4  “ 198.99  inches. 

E 

The  deflection  diagram  of  the  structure  due  to  a moment  of 
one  inch  pound  at  B may  now  he  drawn  to  scale  (Fig. 19) . The  resist- 
ing moment  at  the  base  of  the  right-hand  column,  due  to  a vertical 
load  of  one  pound  at  any  panel  point  of  the  truss,  may  now  be  deter- 
mined. As  before  stated,  this  resisting  moment  is  the  vertical  de- 
flection of  the  panel  point  loaded,  times  the  length  of  the  lever 
arm  of  the  force  which  produces  the  moment  of  one  inch  pound  at  B, 
divided  by  the  deflection  of  the  end  of  the  lever  due  to  the  moment 
of  one  inch  pound  at  B.  The  vertical  deflections  of  the  panel 
points  are  obtained  by  scaling  from  the  Williot-Mohr  diagram.  The 
resisting  moments  at  B due  to  unit  vertical  loads  applied  at  various 
panel  points  of  the  truss  follow: 

1 lb.  vertical  at  3:  =-  ^q'I^qq  " 5.925  in. lbs. 

1 lb.  vertical  at  7:  =■  » 5.292  in. lbs. 

1 lb.  vertical  at  4:  M*  = z 6.196  in. lbs. 

^ 198.99 

The  resisting  moment  at  the  base  of  the  left-hand  col'umn  for 
each  case  is  easily  obtained  by  statics.  Values  of  the  resisting 
moments  at  the  bases  of  both  columns^ due  to  unit  vertical  loads  ap- 
plied at  various  panel  points  of  the  truss,  are  tabulated  in  Table 
IV. 


III.  APPLICATION  TO  THE  DETERLIINATION  OF  REACTIONS 
7.  Deterniinat ion  of  Reactions  Due  to  Vertical  Loads, 

Since  snow  loads,  roof  loads,  and  other  uniform  vert- 
ical loads  on  bents  are  usually  given  in  pounds  per  square  foot  of 
horizontal  projection  of  roof  surface,  the  reactions  due  to  a uni- 
form vertical  load  of  one  pound  per  foot  length  of  truss  are  here- 
in determined.  The  reactions  due  to  any  uniform  vertical  load  ar^ 
then,  equal  to  the  distance  between  bents,  times  the  magnitude  of 
the  vertical  load  in  pounds  per  square  foot  of  horizontal  projec  •- 
tion  of  roof  surface,  times  the  reactions  due  to  a vertical  load 
of  one  pound  per  foot  length  of  truss.  The  vertical  loads  are 
considered  to  be  applied  at  the  panel  points  of  the  upper  chord. 
The  load  carried  to  each  panel  point  is  one  half  the  load  on  each 
adjacent  panel. 

The  horizontal  or  vertical  reaction  at  the  base  of  the 
right-hand  column,  due  to  a vertical  load  of  one  pound  per  foot 
length  of  truss,  is  obtained  as  follows: 

Obtain  the  vertical  deflection  of  edch  upper  chord 
panel  point  of  the  truss  and  the  deflection  of  the  point  B (See 
Fig.  9 or  15)  at  the  base  of  the  right-hand  column  in  the  direct  - 
ion  of  the  desired  reaction,  both  deflections  being  due  to  a force 
of  one  pound  applied  at  B.  Compute  the  portion  of  the  total 
load  on  the  tr  uss  carried  to  eacn  panel  point  of  tne  upper  chord. 
Then 

N = rwd^ 

B » 


-o4- 


where  N is  the  reactioa,  horizontal  or  vertical,  as  the  case  may 
B 

be,  at  the  base  of  the  right-hand  column,  due  to  the  load  of  one 
iiound  per  foot  length  of  truss;  d^  is  the  vertical  deflection  of 
any  upper  chord  panel  point  x of  the  truss;  W is  the  portion  of 
the  total  vertical  load  carried  at  that  panel  point;  and  d is  the  t 
displacement  of  B,  the  base  of  the  right-hand  column,  in  the  di- 
rection of  the  desired  reaction,  i.e.,  horizontal  or  vertical. 

The  resisting  moment  at  the  base  of  the  right-hand 
column,  due  to  a uniform  vertical  load  of  one  pound  per  foot 
length  of  truss,  is  obtained  as  follows: 

Obtain  the  vertical  deflection  of  each  upper  chord  pajiel 
point  of  the  truss  and  the  deflection  at  the  end  of  the  lever, 
where  the  force  causing  the  moment  of  one  inch  pound  at  B is  ap  — 
pliedr^^QQ  Fig.  18),  both  deflections  being  due  to  the  moment  of  one 
inch  pound  acting  at  B.  Compute  the  portion  of  the  total  vertical 
load  carried  to  each  panel  point  of  the  upper  chord.  Then 

M = k Zwd^ 

B 2L.  , 

where  M is  the  desired  resisting  moment,  W is  the  load  at  any 
panel  point  x of  the  upper  chord,  d is  the  vertical  deflection 
of  that  panel  point  due  to  a moment  of  one  inch  pound  acting  at 
B,  k is  the  lever  aim  of  the  force  which  produces  the  moment  of 
one  inch  pound  at  B,  and  d^  + is  the  displacement  at  the  end  of 
the  lever  due  to  the  moment  of  one  inch  pound  at  B. 

The  vertical  deflections  of  the  panel  points  of  the 
upper  chord  are  due  only  to  the  deformation  of  the  truss.  These 
deflections  were  obtained  by  scaling  their  values  from  the  Williot 


• Ttif  I-  . 


?">i  :'^1wm’-.A 


■ '»' ' 


KT.'^/’  • * ''^''  ' - '■'■  ■•  rr^  . ''V,  .J  S • '''i^’-’V^ 

fWB'CJ  ' i .^^  '>J  5''^  U *^«l7'  to  <5eAjBf  P -« 


^ , ''rr  "‘  11:^ 

all*^  fci  ^t)  •«3jlT;^  >,c.‘  ^oolt 


pi.  ;:4"  >• .' .-.j;'"' jiai H'jsi^‘‘  ir X JiT' <# ' tyioJL", 


4^ 


-'  - s»'  ' ‘ ' ' ' ' ■ '“  ' ’•'•'•  t • 'J*!  * 


• ? ' ’•  ?•»  '»-.»,  I <'L:3 

■ .>r*  JUt.K^li^^  , * .1  eflla  Jo  <^4403(111 

■tie  ^ P''  ‘ ' ■'.  - 'I 

fifi.,  4-J'ttoi,V‘’’e^-'  i>.^JfO  f'U  'ad^ ''':'39HI(5^  ' 

A ” . ^.  •/  -V  , ■ ' '- ' \ ^ LL ' + li^ 'i^  “' 

'■  ^••'*0  t..  ,$®ol  jfiaVvtnu i^jpi  ,(mUlco^ 


!►  ‘^SOi  t^TociV 

,‘j,i»*f  ! ^0  ..fen<i' It*’ :T'o.i:J^■tl^s 


,i '■.  ■'  ‘.r  •'  /•^■‘ 

tt»  1 o . iao  id  X1« 0 ' lifox d T.<>y ' o i' d fiX*  it? 6^  ,. 

:.'  ..  .:  -1  ,v  . / ■■'^ . *'  X**  yi  ' ■ 


a 


','Acdv*,0'ei«;d; 


to 

,.  .,4  ' V '-"  i!'--^'^  -'''2!®^'  yr'T 

*<W  -X  ?’l  *.  iii  Liofii  .oAO  ts-  w»o:r  orf/-' %ri:ii»0/,,Tt?50T,^^ 

I . -■  ’‘"“'\  , J‘  . ■*  ' > ' kH^ 


M'*’ ■iO' •V'*"  • V 04  tHja  ttirlatf  •^Q  f5o5)  i'tafc  d^icf  «{6i!  lii 

-irdl’  f^3  adx/.^oO^  :^u£o^p^ftc;t;pq:xJo^i  1:^ 


.'i'. 


r«^f'  f ■■  ■ 

m^''T  *i  4 <(j^  liT'  # 70'  .4 dl  aa , ia w,q , ftp^  1 . W- , b«i  2 «td,o  t^sf,l 

r ; ■ ■':■■' -V'  '^  ■ 

, . . 7~^~  -^■.' 

Ik.'*-  ft'  **  -V*' 

W ■ - .;  -a  - - . *•  . 


. !r^,^  ’.V  • i 


Bb 


■i,y 


'J41 


]0 


’■  ' *■  ''  ■'*'  ' 1 ' t‘  'T'  t»  ' X>?”  “■  -,;  '"  ■ — I ' I 

Mi  ,Mo4»  0<i? 


I',  ''tU  '-e'lOv  to,.  iaioq>Xo/iPd  iS 

■•.i  'S'.;:,.  .—  'y^, 

.itijCfi'-  li’yei  i\4B'':.  i£;| 

1 '.^.  . T8  „ *•',  ■ . '- . ' '>.  . .■  ...  '•  '’”  ;-;■  . f 1 

rt*  ^ iyd'iiti:  ,S  4;d'''^bA:Joa'.dO0Jt  aorf  il 

. .■»  , \ ' ■ ..*3"  ')3?  :,v.S 


> 


if  'te 


.,  >nd  i’JdX^qq  ‘Itiiai;  *dT 

|idlrii'<^T'  .wvtAi  «>rti  to-idG7ioi^*.TOt^ii'!#lF4  ,,W  >i.a6  ax#?Jb«o40‘ 

:v  . X ;!  7*.,  f-i£  ':  ' 

CS.-.'®  X--“.''  'V  -^A«  : ■ ■ ‘■■'.sv^i'-  ■ ■ ■'  . " ' 'lii 

KLiXiLu^  ^ _ -L  Si.  ’ ’‘^  * ■ 

yys  iv??  ^B.c3i>jpi.i|,i';.|)|«,'i‘i,i,^ji  Stesrt  rrgjyr^ifat.^ 


-o5- 


diagrams,  drawn  by  Mr.  S.  R.  Offutt  * for  his  thesis.  The  vertical 
and  horizontal  displacements  at  the  base  of  the  right-hand  column^ 
and  the  deflection  at  the  end  of  the  lever,  where  the  force  caus- 
ing unit  moment  at  B is  applied,  were  also  obtained  from  his 
thesis  for  each  bent  analyzed.  These  deflections  were  originally 
obtained  by  computation  by  the  method  given  on  pages  to 
of  this  thesis. 

The  horizontal  reaction,  the  vertical  reaction,  and  the 
resisting  moment  on  the  left-hand  column,  due  to  a uniform 
vertical  load  of  one  pound  per  foot  length  of  truss,  are  equal 
respectively  to  the  horizontal  reaction,  the  vertical  reaction, 
and  the  resisting  moment  on  the  right-hand  column  for  each  bent 
analyzed;  for  the  load  is  symmetrical  on  the  two  halves  of  the 
truss. 

The  distance  from  the  base  of  either  column  to  the 

point  of  contraf lexure  of  the  column  is  found  from  the  equation, 

M 

M - HY  = 0 or  Y = » 

where  M is  the  resisting  moment  at  the  base  of  the  column,  H is  the 
horizontal  reaction  at  the  base  of  the  column,  and  Y is  the  distance 
from  the  base  of  the  column  to  the  point  of  contraf  lexure  of  the 
column. 

Concentrated  loads  due  to  cranes  sometimes  come  at  the 
panel  points  of  the  lower  chord  of  a truss.  For  this  reason,  some 
panel  point  was  chosen,  and  the  reactions  at  the  base  of  each 
column,  du. e to  a concentrated  load  of  one  pound  at  that  panel  point, 
were  computed.  This  was  done  for  each  lower  chord  panel  point 

of  the  truss.  These  panel  points  are  points  1 and  4 
% 

See  footnote,  page  3. 


'.HI 


^ \r.,i .; • ^.<»t' ' > • i^rv f't  49'i^'•’Jo’VA>•.4'  ^ ii^'-  ^.<s'^-'flt:.'*‘^V ^ ^ |i>  .iu  .toti '-^t^' 

. ' '■  ■ '/■'■  "W  ' 

r ;.*  '-■%:>i  4/-'  .i^mu  ■ ;U0  ‘jayi  ^ 

* ' ^ V«».'  *1*  * '* 


♦ , 


■•■  ^ , , '•’  - '•  ,*i , . , • ■ ,,  ,'::'ii. ■»'!•:•. 


^ --V  ...  v;.  - ■■,  ^ A •'» 


[4,  OS>\:^--  wu;;/ 

f ■ -r  .!• 


'•  /.  i.**' . “■ ' • 


p^  V'  s Ai  '®. 


•p;r' 


,y>  4.>*/  Xj.-  l^nev  '' 


.,  J 


* , ' f ■ ^'  ‘ ^ * ' ” 

ix a,«CH  s-ii  ' =^:av  To,  1»oX' 


•'»  ' ‘^r:.-/tK  s ' . ■»■  ‘ ' ^ -i*yy  ^ , 


: : 


r.  'Pili  To ctii  u oia^&58MTfj|  ri 


i?-‘rij'.  sY-'^ 

^ » , ■ • „ i» 


. - •’ 
■ ‘ *■ 


^v’."iiii 


5 


7/% 


'^ioo  1 -'MH 


' si; r , - ...rjo  a%.Y«‘-K''l 


:V- 


' ■■'.'■Al 


-ii6- 

for  the  bents  having  30-f  t. , ^0-f  t,50-f  t, , and  60-ft.  spans  (See 
respectively  Figs.  23,24,25, and  26  ),  For  the  bent  having  the 
20  -ft.  span  the  point  is  point  T (See  Fig.  22).  The  reactions 
on  the  right-hand  column  are  obtained  in  the  same  manner  as  the 
reactions  due  to  the  uniform  vertical  load,  and  the  reactions  on 
the  left-hand  column  are  then  found  by  the  equations  of  statics,  j 
The  points  of  contraf lexure  for  both  columns  are  obtained  in  the  j 
same  way  as  the  points  of  contraf  lexure  due  to  the  uniform  vertical! 
load.  The  reactions,  due  to  any  concentrated  load  at  any  panel 
point  of  the  lower  chord^  are  directly  proportional  to  the  reactions 
due  to  a one  pound  load  placed  at  the  panel  point. 

8.  Determination  of  Reactions  Due  to  Combined  Wind 
and  Vertical  Loads. 

The  horizontal  react  ion,  in  the  case  of  the  hinged 
columns,  and  the  horizontal  reaction,  resisting  moment,  and  point 
of  contraf  lexure,  in  the  ca^e  of  the  fixed  columns,  due  to  a wind 
load  of  20  pounds  per  square  foot  of  vertical  projection  of  bent 
combined  with  various  uniform  vertical  loads,  varying  from  30  lbs. 
to  100  lbs.  per  square  foot  of  horizontal  projection  of  roof  sur- 
face^ acting  on  the  bent^were  obtained  each  column.  A wind 

pressure  of  20  lbs.  per  square  foot  is  a commonly  assumed  value 
in  this  region.  It  is  the  value  for  which  the  bents  were  designed. 
The  wind  was  considered  to  come  from  the  left,  thus,  making  the 
right-hand  column  the  leeward  column  and  the  left-hand  column  the 


windward  column. 


* t a ^ M t.  ' 

X—”  , ■ ■■'■  ■ , l!.’s 

> ' * J C"  '•■  .'t : -J.  <■ ' ^ »'2S  i ^'t;' . *,  y . -.•..I  ■ ■'  ' ■ ! ' j 


‘..fr3S 


•.i^;,^rl>i'''j'it  . ("^j:'’  -V'^^  fc  sE)  T jiiictfl  •*  jiiiofl  A*t.^ -.AltV  0^ 

.'  • ■••™  • ■ ' i ^1 

« f fU  4,t  bi*iiJi;i.’j?tJv  di*  t t? 


i‘  t\Oi 


j f Id V'lt  Jtte rf#'  *»xa ^ 

■ -■  . % ' V.  ■ . ■ ■ if* 

fti  iw  fcosuii^o- 

pH’ ‘ *Ti>r*tni/ ■•  ■■■*  !^o  -■ua' ^ 


r 


'■  , ^ • • *Si 


lyti^  ■!iiinti%:.hji’Ci:l  batjo^nao.  M^qi 

‘ ^ >.  r >*{(«'•* 


'Jr.;-  >' 


. ■•«  .'  ' 1JH'...'- % wifi 


to  »fip  ff?  .fiOiAWi'  ■ * T 


r‘i<C><5  «<AA  , .tr^apCT'  l'^  f*'^\-f.!(:‘^  I ‘^^'4'T'  •, 


v»  * 


' *■  ."  ■•r;/  , . "'  ’ ' ' ■'  ’ . ■ . J"  ^ ''•^.' 


^-TIS 

■ X ' 


1 . 


G)B 


<sac4«  ePx  aa^, 

If  '•  ^ i ■ , .♦  '"'’*r,I» 

•' ‘*''  'C  ' 4^  » : % j 


^64^f:Jr»^k^  ■ ^ l PlAUPp 

^>'2d  ^fi4  il.E»i.^j|r*’X0'i'  .iiAt2^7  Ml  IN:  ij 


.,,  if.  ,s)  * '‘Att  .^1 


^>'  . ■ '“  ■''* 

pv^  e„dT 


ii.  • *•.  ■ ' ' . , iKf'4 

■^^Mcsa 


-37- 


Values  of  the  horizontal  reaction  and  moment  at  the  base 
of  each  column,  due  to  a wind  load  of  one  pound  per  linear  foot  act- 
ing on  the  windward  column  and  0.7455  pound  per  linear  foot  acting 
on  the  windward  slope  of  the  roof,  were  obtained  from  Offutt' s 
thesis.*  This  horizontal  reaction  or  moment  acting  on  either  columr^ 
times  20,  times  the  distance  between  bents  (15  ft.  for  all  of  the 
bents  analyzed),  gives  the  horizontal  reaction  or  moment  at  the  base 
of  the  column,  due  to  a wind  pressure  of  20  lbs.  per  square  foot. 

The  horizontal  reaction  or  moment  at  the  base  of  either  column,  due 
to  a uniform  vertical  load  of  one  pound  per  foot  length  of  truss, 
times  the  intensity  of  the  vertical  load  in  pounds  per  square  foot 
of  horizontal  projection  of  roof  surface,  times  the  distance  between 
bents,  gives  the  horizontal  reaction  or  moment  at  the  column  base 


The  directions  of  the  horizontal  reactions  and  moments  at 
the  bases  of  both  Inward  and  windward  columns  are  shown  in  the  above 
figure.  is  the  horizontal  reaction  due 'to  the  vertical  loads, 

H is  the  horizontal  reaction  due  to  the  wind  loads,  M is  the 
moment  due  to  the  vertical  loads,  and  M is  the  moment  due  to  the 
* See  footnote  on  p-ge^. 


^vr^„f^  • " '.‘l'.v.  ■.  • ...ii  ■ 


J 


.vr,i.  ' :>:'l^-  'TfOu.  ..■'  J?4  } Pttu.^f,.r^ 


M. 


1^  >:^a4ia,-Xi^  ari^ 

^ ''  ; .1' ,*T^  -V  '•  \C  .if'"-  :®  >* 

•‘  4 . Av^  L '^j  '5^  - ^34  < j>,.’^.'  o’ti^"- ' « tt.i» jcri oiCi.;|J4C 

*'  f ''  ■'’^ 


/■  ’ ■'  ' 7^ 


1 ■/*.;' 


|:  :■;  ^ *«■  ^ l.^-^f.  i-.»x  ft  c ' 10 

• ' *-  4'."  • ' *ti  '•A- 


F^l  . > 

’ J .to’ 


7 7 ij  ' I -ttK  •*  I:i^  ^.:-rA  ■ 


t _i 

i. 


\r-^m^ 


■atjR'14't'  o*i'^';  .‘^yW  ,iycU/4^^'X’  '/.if aai 


~".  ( w "?/' 


Ilk 


‘4/sJ^i'  > ;wo.y.j^jAga 
* Mimv  Ji  4 


'<.  //// 'l! 

^Sk,  fli  ‘ I'k 


4^ 


^,- 


'‘'v 


i 


£-il 


- tM  '• 

J'  ; 

4- 

er’V  ;.a 


ilv, 


V 


';,i.:^',  -:  V-  rm,.  ;:.  , V 

Uj  ^7..4:-.."  *ff  /vi*-. 

' K-  '4  . ■ rV,  ■ ■•  ;■  ■'.  . .. , ' ■ 1t^V'  ' 

'fr^0i::  «;t?' 


^'.ij  _ , 

_ , i 

*' ' ' . ‘ ' ■*  ' ' '‘lB^>f'  ■‘/^1,-1  ,i.  / *^’v,  ' .s'‘.'’Sb  f ij 

, |«oui^jir^4l- '’^9  ttfn 

^ >4  ■ » i^i* Philip  -ttcf  fi^Wi'i 'V-.ft 


y:  . . « j:. 


if>' 

i 4\i 


SI 


V ;:  -xSit pi  ■ ^, 


,'r  ■ rt.-f'  , 


tw#«"?'iiigif'' 


**  y s^qtai^apitera 


-o8- 


wind  loads.  The  combined  horizontal  reaction  on  the  leeward 
column  is  the  sum  of  the  wind  load  reaction  and  the  vertical  load 
reaction  on  that  column.  The  combined  reaction  on  the  windward 
column  is  the  difference  of  the  wind  load  reaction  and  the  ver  - 
tical  load  reaction  on  that  column.  This  same  statement  is  true 
for  the  combined  moment  on  each  column.  Reactions  to  the  left 
and  counterclockwise  moments  are  considered  to  be  positive;  re- 
actions to  the  right  and  clockwise  moments  are  considered  to  be 
negative.  The  combined  reaction  on  the  leeward  column  is  always 
positive,  and  the  combined  reaction  on  the  windward  column  is 
positive  or  negative,  depending  upon  whether  the  reaction  due  to 
wind  loads  or  the  reaction  due  to  vertical  loads  is  larger. 

The  location  of  the  point  of  contraf lesure  on  the  leeward 
column  is  obtained  from  the  equation. 


where  Y is  the  distance  from  the  base  of  the  leeward  column  to 
R 

the  point  of  contraf lexure  in  inches,  M is  combined  wind 

R 

load  and  vertical  load  resisting  moment  at  the  base  of  the  column 

in  inch  pounds,  and  H is  the  combined  wind  load  and  vertical 

R 

load  horizontar' react  ion  at.  the  base  of  the  column  in  pounds. 

For  the  windward  column. 


LL  ~ ^ Y = e or  Y. 
R R R 


R 


or 


12 


va  'i'l  , wf^S/lf^  * * * P.\  '<k,. 


' '■*  'A-  ' ’ ' ■ 


V 


iV 


)Cf.^ 


'“FjS’  '’f£i 

bciillV' 


tf  C'X  _f*:j>{'l4']L'<>J;^  „.ti^  W;.ia  -iSCi  ti^O £» ,|'!tl,tiVvo  •m  <*oros 

. ' , ■.•'■*  3b..  ^it  --.^.,-' 


U^. 


• 'j4 'i' s: • i.  4©-*  l^ai' 


« Ik:  w '>iis  *T’ft  flf-  ♦ i fi-Ht  4;4-  ••; 

rt-  . ■ . ,‘V.  ' .te.  ■ .,: 


; ■ y JW  i 


>iwij  ' ■ .**  v'' 

i^BV5  •./■_  <l.'.^  > *.  B f .4,^  v^>*i  » tifci  .'^  *!■  ''  rtJr.t,  "♦ 'jJsfikfc''  ^ » i-/  '. 'Sfl 


ST 


* , jsl: .-’.  * ar  'J « ■ 4c  p '.;>-■ 


% 


5J 


..‘•as? ; *^:,'*;  V4V' - fUK 


■'^^1  ^ 


y » 

-,  U ' far  Us  ■ 'i?'  :‘«o  jrx>  ?>■  0-  6k'  '^>:C>XVi 

■ , • ^ 


'^y  ' “ 


* , - ■ '^‘■'^>  ^ .'1''^ ,.  •.‘^^4ii™i 

. i!:"?!  ff  t 4 it>^  *.  . . « ^ . --i  ,.  . ^ 


jf‘it\.,a  v*5'  ‘ r: 

it 


’■  ' ' ’ i iiv'  ' k Hi  - ' -'  '•  ’ Ti'/  iK 

■a*  . " f ^ lif  , ,^»  «+\c*ii^  aV  oi^^t  ^ 

' ■ ff  ‘ ■ ''  :'  'i.  ■ F' 

ic  *n\H;}  . ^/j  .+.•  ir<  .-)g^«  . •><>/, u^.«i-Jr  f^ 


k*..'  .; 


JA’ 


'X- 


/ivi'  .ftiv'  Xs- .M.  ■■  ^.  ^ 


n« 


-#'4 


• ,'Vv 


^v,r€  niiifjti^piH 


* K.''‘*l  '*' 


'/a  ':^  .-i 

' ^ ’ 1.  ► ^ 7T)-V.t 


Jf#< 


Pliri.*  !^  ;.-■: 

MP^rref^ 

l^r. 


<y*  , yy  \ 

' '.'  • lliirrlfi™®™"' 

''■'i*'" 
im 


f ‘ 


SRWBtie  f samT-rarr, 

/j  • ' •*  ..  ■ ' r 


-o9~ 

where  Y is  the  distance  from  the  base  of  the  windward  column  to 
L 

the  point  of  contraf lexure  in  inches,  H is  the  combined  wind  load 

L 

and  vertical  load  horizontal  reaction  at  the  base  of  the  column 
in  pounds.  is  the  combined  wind  load  and  vertical  load  re  - 
sis  ting  moment  at  the  base  of  the  column  in  inch  pounds,  and  W is 
the  wind  load  in  pounds  per  linear  foot  on  the  vJindward  column 
(20  X 15  = 300  lbs.  for  the  wind  pressure  assumed). 

9.  Bents  Analyzed. 

The  results  of  the  analysis  of  twenty  mill  build~ 
ing  bents  the  type  shown  in  Fig.  1,  consisting  of  one  quarter 
pitoh  Finn  trusses  mounted  on  columns,  are  presented  in  this  thesis 
Line  diagrams  of  the  bents,  drawn  to  the  same  scale  with  dimen- 
sions, are  shown  in  Fig.  21.  Each  truss  has  eight  panels,  except 
the  trusses  having  the  20-ft.  span,  which  have  four  panels.  The 
trusses  having  30-ft.,  40-ft^  cuid  ^0-it.  spans  were  designed  for 
a vertical  load  of  about  40  lbs.  per  square  foot  of  horizontal 
proj  ect ion'-of  roof  surface^  considering  the  bents  placed  fifteen 
feet  apart.  The  trusses  having  the  2G-ft.  span  were  made  much 
heavier  than  necessary  to  see  what  effect  this  might  have  on  the 
results.  The  truss  design  is  constant  for  each  span  length. 

The  columns  of  the  bents  having  the  30-ft.,  40-ft., 
50-ft  , and  SO-ft.  spans,  with  the  16-ft.,  2.1-ft.,  and  26-ft. 
column  heights,  have  the  same  section.  This  design  was  for  the 
40-ft.  span,  21-ft.  column  height,  assuming  a wind  pressure  of 
20  lbs.  per  square  foot  of  vertical  projection  of  bent,  and 
assuming  that  the  horizontal  reactions  were  equal,  and  that  the 
location  of  the  point  of  oontraf lexure  in  each  column  was  at  a 


’-^1 

■ ,» 


1 

'•[  . 


4ai4x0i-. 


; *. . ' ■•'.  i-‘ 

Sv 


-i 


; .1 


>v*'£ • ri.  f.-.t.  C 
•,  i.ii  ..  X 

: : ■ .;’  '’  ;■ 


: x :.  - 


to  Jil. 

. . V L.T' 


..  ■ihki  -i:':'  ■’(  ■ ;'^-i  . ■•■  ' i:  .i>.  - ^ ^ 


lV-  ^ 


fi  ' 


' * 

oV-‘.->  . 


'*.  I 1 ‘ 

.'  '•  ^ '.'/iv;'*  ' !'.' 


t ^ 6'\] 


^4 


■ ■ f f 


il  ,y 


f**' 


' y ’lJO,’ 


• :•  ..■  fr.-:.  ..'■  . 

,Z' ' ’■-■ . ■ I >'  . . , 

r 

'\  vXu  V' 

•.Oi'.'"'  ; .IOC 

■ • V • 

: . ■ ..  .•  c 


;,.:  U;  .,vC^  on  ; 

„ ' • , ;'• '),  ..  ■ 

if  ;:i-  ■ ; •■:  t >t«  ■•  - 


4? 

/. 


) ' ' 


:.:a 


■ 


.W 


■ 9k-y, 


..  ./ . ',t- 'i;-  -u  ....  •-,1  . - . 

..  i:  ■•>.  1,3  . . .;■■  ■“•  ' ni.-v.' 


> I 


.i  '■■•• 


.'•.  1.; 


.'  O'-;  .! 


.»'• 


'i  ’A  oi,:^ •;•-’*•  1'-^  ,1  p.,  !i)9f« 

• ' > ,1  ‘ * - 'V.  ' '■  :•  • 


iyzy.m.  .>  J 


■Ml.  .< 


JL.»uV'  •'■  A 


■^  /i','&  -ft'  f ■ >r.  Ci.  - ; ••  1. 


'T:rye»&. 


-40- 


pointy  one  third  the  distance  from  the  base  of  the  column  to  the 
foot  of  the  knee  brace.  The  columns  which  are  31  feet  high  eere 
made  considerably  heavier  than  necessary  to  see  what  effect  stiff 
columns  would  have  on  the  results.  The  columns  of  the  twenty-foot 
span,  which  are  10-ft,,  12-ft.,  and  16-ft.  high,  have  the  same 
section,  which  is  hfeavier  than  required.  The  19-ft.  column  has 
a lighter  section. 

The  bents  having  the  2©-ft.  and  30-ft.  spans  have  a 
shorter  distance  between  the  foot  of  the  knee  brace  and  the  top 
of  the  column  than  the  other  bents.  This  distance  was  selected 
to  see  what  effect  the  position  of  the  knee  brace  waald  have  on 
the  results. 

Figures  22  to  26  are  line  diagrams  of  one  bent  of  each 
span  length.  Points  J,  G,  F and  E are  the  girt  points  of  the 
column  DJ.  Tables  1,5, 6, 7,  and  8 give  the  designs  of  the  bents. 

10.  Results. 

A complete  analysis  of  one  bent  is  given  on  pages 
<o  to  '^'2-of  this  thesis.  Figures  1 to  20  and  tables  3, 3,^4  are 
given  to  accompany  this  analysis.  The  Williot  diagrams,  from 
which  the  vertical  deflections  of  the  panel  points  of  the  truss 
were  scaled,  are  given  only  for  the  bent  havihg  the  40-ft.  span 
and  21-ft.  column  height.  These  deflections,  together  with  the 
calculations  necessary  to  determine  the  reactions  due  to  vertical 
loads,  are  given  on  pages 

In  order  to  compare  the  values  of  the  reactions  due 
to  vertical  loads  for  the  different  bents  , a comparison  factor, 
which  is  the  value  of  the  reaction  divided  by  the  total  height  of 

pj 

the  bent,  was  chosen.  Thus,  '"L  is  the  comparison  factor  for  the 

t~~ 


>>f 


■ ■ ..Y'tt;- 
f'.'j 


\ 


..'7  < ' ' 


'u  ^ 


■'■'Vy  .,.  • ■.  i’-'-JOc  vi.:  > V'-^  ■■  .:iv 

*!*^  . . ' ' 

'pi/ c *i'  , ■ r.  t ■:..i,Xb-^.t‘.i.  :>  ^ri:t  #.>i 


t 


-41- 


horlzantal  reaction  at  the  base  of  the  left-hand  column,  where 

H is  the  reaction  and  t is  the  total  height  of  the  bent. 

L 

Table  0 gives  the  comparison  factor  for  each  reaction, 
and  also  the  ratio  X/ d for  each  column,  where  Y is  the  distance 
from  the  base  of  the  column  to  the  point  of  contraf lexure  and  d is 
the  distance  from  the  base  of  the  column  to  the  foot  of  the  knee 
brace,  due  to  a uniform  vertical  load  of  one  pound  per  foot  length 
of  truss.  Tables  10  and  11  give  the  same  quantities  due  to  a con- 
centrated load  of  one  pound  at  same  panel  point  of  the  lower  chord 
of  the  truss. 

Values  of  the  horizontal  reactions^ due  to  a wind  load 
of  20  lbs.  per  square  foot  of  vertical  projection  of  bent,  com  - 
bined  with  various  uniform  vertical  loads,  are  given  in  tables  12 
and  14  for  hinged  and  fixed  columns  respectively.  Tables  13  and 

15  give  the  values  of  the  ratio  R for  hinged  and  fixed  columns 

respectively,  where  H is  the  horizontal  reaction  at  the  base  of 
the  right-hand  or  the  leeward  column  die  to  the  combined  wind  and 
vertical  loading,  and  H is  the  total  horizontal  component  of  the 
wind  load  of  20  lbs.  per  square  foot  acting  on  the  bent.  Table  16 
gives  the  values  of  the  resisting  mom*ents  at  the  bases  of  both 
columns,  and  table  17  gives  the  location  of  the  points  of  contra- 
flexure  on  both  columns,  due  to  the  combined  wind  and  vertical  loads 

Graphs  No.  1 to  6 shovir  the  effect  of  variations  in 

different  characteristics  of  the  bents  on  reactions  due  to  vertical 

loads.  Graphs  No.  7 to  36  show  the  effect  of  these  variations  on 

the  ratio  and  the  location  of  the  point  of  contraf lexure  on 

R 

H 

each  column.  These  graphs  will  be  discussed  later. 


W '^'  4 rit  :\!,i  ,f?kx*£  v*^ . 'TrtK  V'J’  I ^0'  j > if'jBv.  ^Q,l:Jf j6>#«'^  JC4».i' 


•jrrv'raatwTyj 


-I-.,  •®1 


,»  ■ 


V*' 


i',  /^*J 


» - :':ifa  ,t*  -.r<)u#l  /iOk5ttC(inoo ■ etbi^ijj.'V  ■ 

' ' •••  . . ■ . i ' ^ "'iQ 

Ix  L"!!:  "t  '.-lo.  f jJor*..  tot  .'::>  t " r^¥’4jH?  til 


; ;•  'f'Ur  t X ■'v^ : ».  0 jc/.o<3  .r 

, . •!  •'  ;'•  ' ■'  : , ^ ;a  ' .-  ;^ 

■n^iod  '\o  . 


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-42- 

IV.  TEST  OF  PAPER  MODEL  OF  BENT. 

11.  Description  of  Test. 

In  order  to  check  the  reactions  computed  according 
to  the  theory  presented  herein,  a paper  model  of  a bent  was  con- 
structed, and  the  reactions  were  obtained  from  measured  deflections 
of  the  model  and  compared  with  the  calculated  values. 

A model  of  the  bent  having  the  40  - ft.  span  and  21- 
ft,  column  height  was  made  of  heavy  paper.  The  span  of  the  model 

was  15  inches.  The  truss  and  columns  were  so  proportioned  that  the 

moment  of  inertia  of  the  truss  of  the  model  _ the  moment  of  in- 
mOrnent  of  inertia  of  the  steel  truss  the  moment  of  in- 

ertia of  the  column  of  the  model  . This  method  of  proportioning 
ertia  of  the  steel  column 

the  truss  and  columnswas  selected,  because  the  truss  acts  as  a beam 
supported  on  the  two  col\imns.  In  getting  the  moment  of  inertia  of 

the  steel  truss  or  the  truss  of  the  model,  a section  was  cut  by  a 
plane  normal  to  the  lower  chord  of  the  truss  at  the  quarter  point 
of  the  truss.  Any  other  point  might  have  been  chosen,  provided 
the  same  point  were  taken  on  both  model  and  steel  truss.  The  weA 
members  were  neglected,  and  the  areas  of  the  chord  members  were  con- 
sidered as  concentrated  at  the  centers  of  gravity  of  the  chords. 

The  gravity  axis  of  the  section  in  a plane  at  right  angles  to  the 
plane  of  the  truss  was  found.  Then,  the  moment  of  inertia  of  the 
section  of  the  truss  was  the  area  of  each  chord  member  times  the 
square  of  the  distance  from  the  center  of  gravity  of  the  chord  to 
the  gravity  axis.  The  truss  members,  themselves,  vvere  proportioned 
according  to  the  areas  of  the  sections  of  the  members.  A drawing 
of  the  model  is  shown  in  Fig.  27. 


-4o- 


The  model  was  floated  on  small  steel  bearings  placed 
on  a drawing  board.  A small  weight  was  placed  on  the  model  over 
each  bearing  to  prevent  buckling  of  the  model.  In  getting  the 
deflections  of  the  model,  the  left-hand  column  was  fastened  at  its 
base  to  the  drawing  board  by  needles,  and  deflections  were  applied 
at  the  base  of  the  right-hand  column  by  a micrometer.  In  the  case 
of  the  hinged  column,  a single  needle  was  pressed  through  the  base 
of  the  ISft-hand  column  as  a hinge;  in  the  case  of  the  fixed  column^ 
three  needles  were  used  to  prevent  rotation  of  the  coluJfin, at  its 
base  when  the  deflections  were  applied.  To  obtain  the  deflections 
of  the  panel  points  of  the  model,  a microscope,  containing  cross- 
hairs in  the  eyepiece  and  mounted  on  a micrometer,  was  used.  It 
was  focussed  on  needle  points,  pressed  through  the  model  at  the 
panel  points. 

The  method  of  obtaining  the  horizontal  reaction  at  the 
base  of  the  right-hand  column,  die  to  a load  applied  at  any  panel 
point  of  the  truss  in  any  direction,  will  no?/  be  described  .♦  | 

The  microscope  was  placed  over  the  panel  point  where  the  deflection  j 

j 

was  to  be  measured  and  oriented  so  that  the  motion  of  the  micrometer 
was  parallel  to  the  direction  of  the  desired  deflection. 

* For  a full  discussion  of  the  method  used  see  a paper 
presented  by  Mr.  George  E.  Beggs  before  the  American  Concrete 
Institute  at  Cleveland,  Ohio,  Feb. , 1922,  on  "An  Accurate  Meehan  - 
ical  Solution  of  Statically  Indeterminate  Structures  by  Use  of 
Paper  Models  and  Special  Gages." 


IHi'rAtr  - 


Ki7»'  vl:*' 


'ln'-ii-'i  ■ V'l  »J«  MiJSii  ac'* ii«SStfl'<A r'f 8b*i 

' ' ,.  "'V  ■ -v-4: 


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44 

A cross-hair  was  brought  over  the  needle  point  by  the  micrometer 
and  a reading  of  the  micrometer  was  taken.  A horizontal  deflec- 
tion was  applied  at  the  base  of  the  right-hand  column  by  another 
micrometer,  which  caused  the  needle  point  at  the  panel  point  to 
move  away  from  the  cross-hair  of  the  microscope.  The  cross-hair 
was  again  brought  over  the  needle  point  and  the  micrometer  read. 
The  difference  of  the  two  readings  of  the  micrometer  was  the  de- 
flection of  the  panel  point  of  the  truss.  Then,  from  Maxwell's 
theorem,  it  follows  that 

Hr  = ^ , 

where  is  the  horizontal  reaction  at  the  base  of  the  right-hand 
column,  due  to  a unit  load  applied  at  the  panel  point  x of  the 
model  where  the  deflection  was  measured  and  applied  in  the  di- 
rection of  the  measured  deflection;  is  the  measured  deflection 
of  the  panel  point;  and  dg  is  the  horizontal  displacement  applied 
at  the  base  of  the  right-hand  column. 

The  vertical  deflection  at  the  base  of  the  right-hand 
column  was  found  by  the  same  method.  The  resisting  moment,  due 
to  a unit  force  acting  at  any  panel  point  and  in  any  direction, 
was  obtained  by  applying  a deflection  at  the  end  of  a short  lever, 
which  was  an  extension  of  the  base  of  the  column,  and  by  reading 
the  deflection  of  the  panel  point  of  the  model  as  before.  Then, 
from  Maxwell's  theorem. 


where  Mr  is  the  resisting  moment  at  the  base  of  the  right-hand 
column,  due  to  a unit  load  applied  at  the  panel  point  x of  the 
model  where  the  deflection  was  measured  and  applied  in  the  dir- 


’■'  . *,  'r  ' , . '■  V i '^f-r  i.  ^.-vl  i. '1'  ■ -'p^'i 

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o;  oc,t.3*  asfliox  diSven  i]tOi!uff^ ‘ii$#©.‘tid*'.oi.'ir 

■ - ■'  ‘ ■ 

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i:0A7G*-.OT.>i  ( ^4lJ  ijgl* 

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\ ^ . ’ • -^  IZ  ',  • ■ • ' '.  'iki  , ^ ‘ ^ 'Wr'  -j 

■ ^ ^ fr'  V ^ ^ v’  K V ' T 

x ;ttii’. .[  Xe,-:>.  ■•  ,6id*  ..-.U  'teXtJroj  KeciX 


I : >•'. y.7v*)fc;ioni  aMWv'mxi^:;i 

'iXlt  "‘  •-■  ' ’ *•  . •'  ^ > '7A#3  > - i. 


ii'-*'-  vM  ■ CT«ii<W|  ’TfS^  *4>  aoio^i:tlSb' ^’'W  * 


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jiilltf"4*x,  . ‘«?  r:-,r’  '■*  •'  '*■  ,•  tJU’ 


-45- 


-eetion  of  the  measured  deflection;  k is  the  length  of  the  levee; 
d is  the  measured  deflection  of  the  panel  point  of  the  model;  and 

X 

d is  the  deflection  applied  at  the  end  of  the  lever. 

B 

The  horizontal  reaction  at  the  base  of  the  right-hand 
column,  in  the  case  of  both  hinged  and  fixed  columns,  and  the 
vertical  reaction  and  resisting  moment,  in  the  case  of  the  fixed 
columns,  due  to  unit  loads  applied  at  various  points  of  the  model 
and  in  different  directions,  were  obtained. 

The  measured  values  of  the  reactions,  together  with  the 
computed  values,  are  given  in  table  18. 

12.  Results  of  Test. 

In  general,  the  agreement  between  the  measured  values 
and  the  calculated  values  of  the  reactions  is  reasonably  close 
(See  table  18).  The  calculated  and  computed  values  agree  more 
closely  in  the  case  of  the  vertical  reactions  and  resisting  moments 
than  in  the  case  of  the  horizontal  reactions.  This  closer  agreement 
is  probably  due  to  the  fact  that  the  deflections  of  the  panel  points 
of  the  truss  for  a given  deflection  at  the  column  base  are  much 
smaller  in  the  latter  case  than  in  the  former  cases.  Any  error, 
due  to  the  slip  of  the  micrometer  or  in  setting  the  micrometer, 
would  thus  be  a larger  per  cent  of  the  total  deflection  of  the 
panel  point,  in  the  case  of  the  horizontal  reactions.  This  is 
especially  true  in  determining  the  horizontal  reactions  due  to 
vertical  loads.  In  this  case,  a displacement  at  the  base  of  the 
column  of  0.100  of  an  inch  produced  a deflection  never  larger  than 
two  or  three  thousandts  of  an  inch  at  the  panel  point.  The  measured 
values  of  the  horizontal  reactions,  in  the  case  of  the  hinged  oolumni 
are  , in  general,  somewhat  larger  than  the  calculated  values.  This 


i ' 

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■’  '"''  'it'  ■'  '‘'  "!  .'  ' , ' ■i''"  j . <■'  '',  - ,■  li 

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pstkd'^^i  fil  ' isLjfrdz 


”,  - . ‘l  ’ ,'  J^-  ‘A',,;'  ’'I'lfjL'V'  ^ ^'2  V ■ •!■  ' '. 

Jii«yi,c  c . ei^-3  tP  o' ajt  *;*«««> i .i  IkWPaXlP^v 


-(•  : 


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artsssriawi?^::  ■-:--r=r-3C^ 


-46- 


fact  is  probably  due  to  a slight  horizontal  movement  of  the  pin  use  . 
as  a hinge  at  the  base  of  the  left-hand  column. 

In  the  method  of  calculation  presented  in  this  thesis 
the  truss  is  assumed  to  be  pin-connected  at  both  points  where  it 
is  connected  to  the  columns,  and  the  truss  members  are  assumed  to 
be  pin-connected.  The  model  had  rigidly  connected  members,  and 
both  flexure  and  direct  stress  could  be  transmitted  to  the  truss 
members  at  the  joints.  The  condition  represented  by  the  model 
more  closely  resembles  the  actual  condition  of  the  joints  of  the 
truss.  However,  the  close  agreement  of  the  calculated  and  measured 
reactions  shows  that  the  difference  in  the  condition  of  the  joints 
does  not  materially  affect  the  results.  The  test  as  a whole  support 
the  theory  upon  which  the  method  of  calculating  the  react! onspre  - 
seated  in  this  thesis  is  based. 


'■  ■ ’.>"'  » '}  ® 'It. 


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,.v, 

m 


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>ii\:  ,-iv  -0«»-  '6itt  ’5#-' 

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••  r%  rr  .’4^^''  ^ 


47 


V.  DISCUSSION  Am>  CONCLUSIONS. 

13.  Effect  of  Variations  in  Different  Cliaracteri sties 
of  the  Bent  on  Reactions  Due  to  Vertical  Loads.- 

Horizontal  Reactions 

Graph  1 shows  the  reaction  factors  for  horizontal  reactions^ 
H/t,  due  to  a uniform  vertical  load  of  one  pound  per  foot  length 
of  truss,  plotted  with  column  heights  as  abscissas.  Graph  2 
shows  the  same  reaction  factors  plotted  with  span  lengths  as 
abscissas. 

The  effect  of  variations  in  column  height  on  the  horizontal 
reactions  is  shown  by  Graph  1.  In  general,  as  the  coltcm  height 
decreases,  the  horizontal  reactions  increase.  Between  the  21-ft^ 
and  16-ft.  column  heights  the  increase  is  very  rapid,  in  the  case 
of  both  hinged  and  fixed  columns;  the  reactions  more  than  double 
in  value. 

The  effect  of  the  size  of  the  column  on  the  horizontal  re- 
actions is  shown  by  the  bend  in  the  lower  portion  of  each  curve  of 
Graph  1,  except  the  curve  for  the  bent  having  the  20-ft.  span. 

A much  heavier  column  section  was  used  for  the  bents  having  a 
31-ft.  column  height  than  for  the  bents  having  other  column  heights  , 
The  effect  is  to  increase  considerably  the  horizontal  reactions 
for  all  bents  which  have  the  stiffer  columns.  This  effect  is  also 
shown  by  Graph  2,  where  the  curve  for  the  31-ft.  column  height  liesi 
above  the  curve  for  the  26-ft.  column  height. 

The  effect  of  the  position  of  the  knee  brace  is  shown  by  the 
broken  lines  on  Graph  2.  The  distance  between  the  foot  of  the 
knee  brace  and  the  top  of  the  column  was  made  four  feet  for  the 


’ I 


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t.  .. 


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I' 


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.>  J ..  . U ' i T ■'»  t ^..  t 

. ■ ■;  . ' •('.{>“  : : orcoi 

0 . . ' ■ , . :i  ’•-  ' 


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, •■:  : ■ .:  -T.i'jv  , • 

' e* 

j'.)  '.:r  ©oiu  o''  'id  iv->L 
' r..o'  ) nv'T'  ; 
oj:'/  rf.r.ro--  i 


, : i:/ a'l"'  \:l 
t> ,' . 


r>  1 


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■■Tsrwessr^- 


-4s- 


48 

bents  having  the  20-foot  and  30-foot  spans,  while  it  was  made  six 
feet  for  the  bents  having  other  spans.  The  points  for  bents  having 
the  distance  of  six  feet  between  the  foot  of  the  knee  brace  and  the 
top  of  the  column  fall  on  a smooth  curve  for  each  column  height. 

The  points  for  bents  having  the  20-and  30-foot  spans  fall  consider- 
ably below  these  curves.  The  broken  lines  are  drawn  through  these 
points.  Thus,  the  effect  of  placing  the  foot  of  the  knee  brace 
higher  up  on  the  column  is  to  decrease  the  horizontal  reactions. 

The  effect  of  variations  in  span  length  on  the  horizontal 
reactions  is  shown  by  the  smooth  portion  of  the  curves  on  Graph  2. 
All  of  these  curves^ except  the  curve  for  the  fixed  column  having 
a height  of  16  feet,  are  nearly  horizontal.  This  shows  that  vari- 
ations in  span  length  have  only  a small  effect  on  the  horizontal 
reactions. 

The  degree  of  fixity  of  the  colmin  bases  has  a large  effect 
on  the  horizontal  reactions.  This  is  shown  on  Graphs  1 and  2,  wher< 
the  curves  for  the  fixed  columns  are  always  above  the  curves  for  th( 
hinged  columns.  In  general,  the  values  of  the  horizontal  reactions 
for  any  bent  of  the  type  analyzed,  having  fixed  columns,  are  some- 
what more  than  twice  the  values  of  the  reactions  for  the  bent,  when 
the  columns  are  hinged. 

Refer  to  the  seventh  column  of  Table  XII.  The  value  of  the 
horizontal  reaction  due  to  a vertical  load  of  one  pound  per  foot 
length  of  truss  is  1.514  lbs.,  in  the  case  of  the  bent  having  the 
20-foot  span  and  10-foot  column  height.  This  value  is  about  7.5^ 
of  the  total  vertical  load  on  the  truss.  In  the  case  of  the  bent 
having  the  60-foot  span  and  26-foot  column  height  the  horizontal  re- 
action is  slightly  greater  than  1^  of  the  total  vertical  load  on 
the  truss.  These  two  values  are  respectively  the  largest  and 


49 


smallest  values  of  this  percentage,  in  the  case  of  the  hinged  col- 
umns. In  the  case  of  the  fixed  columns  (see  the  seventh  column 
of  Table  XIV),  the  largest  value  of  this  percentage  is  15.5^^,  this 
value  being  for  the  bent  having  the  20-foot  span  and  IC-foot  column 
height.  The  smallest  value  of  the  percentage  is  for  the  bent  hav- 
ing the  30- foot  span  and  26-foot  column  height  and  is  about  2.5'J^. 

In  at  least  the  case  of  the  shorter  column  heights,  the  horizontal 
reactions  are  sufficiently  large  to  cause  a »iaterial  compressive 
stress  in  the  knee  braces  and  to  affect  materially  the  stresses  in 
the  members  of  the  truss  due  to  vertical  loads.  The  stresses  in 
the  mem^bers  of  the  truss  due  to  vertical  loads  would  be  somewhat  de- 
creased by  taking  into  account  the  deformation  of  the  truss. 

Resisting  Moments . 

Graphs  No. 3 and  4 show  the  reaction  factors  for  resisting 
moments,  due  to  a uniform  vertical  load  of  one  pound  per  foot  lengtl 
of  truss,  plotted  with  columm  heights  and  span  lengths  respectively 
as  abscissas.  Graph  3 shows  that  the  effect  of  stiff  columns  on 
the  resisting  moments  is  even  greater  than  the  effect  on  the  hori- 
zontal reactions.  This  effect  is  shown  by  the  sharper  bends  in  the 
curves  for  moments  between  the  points  for  the  26-foot  and  31-foot 
column  heights.  The  moments,  in  general,  increase  rapidly  as  the 
column  height  decreases.  The  position  of  the  knee  brace  greatly 
affects  the  resisting  moments,  which  effect  is  shown  in  Graph  4 by 
the  portion  of  the  curves  drawn  in  broken  lines.  A shortening  of 
the  distance  between  the  foot  of  the  knee  brace  and  the  top  of  the 
column  decreases  the  resisting  moments.  As  the  span  leng-th  in- 
creases the  values  of  the  resisting  moments  increase. 


ri!'  * V . 


''  r ■ 


<-w« 


(y^H  r"«, 

vVI 

-s:™  .■v” 


■ i-'.j 


,0  c iJ kI 

< > I . "'.,  '■« 

0\  L '•  * *\  t‘  i. 


V jt  ^ y.  !?■ 

^ ft  i.  U >' 


. V 


li 


■:j 

f 1,  r t,.:  -V 


• . ; i .0  , ■ ‘ ■!  r" 

* 

' , V . ^ t ^ ^ \ i - 

- i- J-.‘  ^■■'  *6':^  y* 

'V^o:.  , . V r 

'*e  ' -^’Z  or'  ■ 


. V/ 

i ;.i_i.'.^ ,.  ^ . 


. i 


'.u'o  . t! <( 


■ov  rf  t I 


' r 

o '.)  • i' . vj;  ; '^''^1... 

’■  *■  - V k r „ 

;.  .....  ...^  .-  l.^tL  , • 


r£!s%.c  .:.’•  iiX  ’i 


1 


I'rS-’j  /■\:r  tyior  :v:v 


a-*..  ,..^.  ..'  *t 


i ■•  V -s. 


50 

The  horizontal  reactions  and  moments  are  thus  affected  by- 
variations  in  span  length,  in  column  height,  in  size  of  column^  and 
in  position  of  knee  brace.  Because  of  the  effect  of  variations  in 
so  many  characteristics  of  the  bent  on  the  reactions  due  to  verti- 
cal loads,  it  was  impossible  to  present  empirical  curves  for  ob- 
taining the  reactions  due  to  combined  wind  and  vertical  loads,  as 
the  writer  had  hoped  to  do. 

Vertical  Reactions. 

Graphs  5 and  6 give  the  reaction  factors  for  vertical  re- 
actions, due  to  a concentrated  load  of  one  pound  at  panel  point  4 
of  the  lower  chord,  plotted  with  column  heights  and  span  lengths 
respectively  as  abscissas.  These  curves  are  approximately  parallel 
and  consistently  spaced,  showing  that  column  height  and  span  length 
are  the  only  characteristics  of  the  bent  materially  affecting  the 
reaction  factors  due  to  vertical  loads.  The  vertical  reactions, 
themselves,  are  dependent  only  on  the  span  length.  The  degree  of 
fixity  of  the  column  bases  has  little  effect  on  the  vertical  reac- 
tions. This  fact  is  shov/n  by  the  approximate  coincidence  of  the 
curves  for  hinged  and  fixed  columns  on  Graphs  5 and  6. 

14.  Errors  in  Comii'.on  Assumiptions  for  Various  Combinations 
of  Wind  and  Vertical  Loads. 

AssuiBption  of  Equal  Horizontal  Reactions. 

The  ratio,  5^:  , is  the  ratio  of  the  horizontal  reaction  on 
H 

the  leeward  column  to  the  tots.l  horizontal  component  of  the  wind 

acting  on  the  bent.  Graphs  7 to  11  show  the  ratios  — ^ due  to  va- 

H 

rious  combinations  of  wind  and  vertical  loads,  plotted  with  column 
heights  as  abscissas.  Graphs  12  to  16  show  the  same  ratios  plotted 
with  span  lengths  as  abscissas.  Graphs  17  to  21  show  these  ratios 

I 

I 

g— _ _ ■ ■ ■ ■ . . — — 1| 


O- 


51 


plotted  with  the  ratios  ag  abscissas. 

span  length 


Hr 


The  first  set  of  graphs  shows  that  the  ratio  — £1  is  large/' 

H 

for  the  shorter  columns  having  the  16-foot  and  21- foot  heights  than 
for  the  longer  columns  having  the  26-foot  and  31-foot  heights.  Any 
vertical  load  tends  to  increase  the  value  of  the  horizontal  re- 
action on  the  leeward  column  over  what  the  value  of  the  reaction 
would  be  due  to  wind  loads  alone,  and  to  decrease  the  value  of  the 
horizontal  reaction  on  the  windward  column.  The  larger  the  value 
of  the  vertical  load  is^the  larger  will  be  the  value  of  the  ratio 

_H  . In  the  case  of  the  hinged  columns,  this  ratio  becomes  nearly 
H 

equal  to  unity,  in  the  case  of  the  16-foot  column  height,  but  never 

exceeds  one  (See  Graph  Ro.ll).  In  the  case  of  the  fixed  columns, 

under  the  vertical  loads  of  60,  75,  and  100  lbs.  per  square  foot, 

the  horizontal  reaction  on  the  windward  coluiian  for  the  short  column 

heights  changes  direction.  It  acts  to  the  right  instead  of  to  the 

left,  and  the  ratio  & becomes  greater  than  unity.  This  change  is 

H 

shown  on  Graphs  9,  10, and  11. 

Hr 

Graphs  17  to  21  show  that  the  ratioj||i^  decrease  as  the  ra- 
tios - Hpi-Sy  increase.  The  ratio  the  case  of  the 

span  length 

hinged  columns,  ranges  from  a minimum  value  of  0.36,  when  the  ver- 
tical load  is  30  lbs.  per  square  foot  and  the  ratio  of^  is  equal 
to  1.033,  to  a maximum  value  of  0.95,  when  the  vertical  load  is  100 
lbs.  per  square  foot  and  the  ratio  of?L  is  equal  to  0.500.  In  the 
case  of  the  fixed  columns,  this  ratio  ranges  from  a minimum  value 
of  0.33,  when  the  vertical  load  is  30  lbs.  per  square  foot  and  the 
ratio  cfTL  is  equal  to  1.033,  to  a maximum  value  of  1.62,  when  the 
vertical  load  is  100  lbs.  per  square  foot  and  the  ratio  ofl^L  is 

equal  to  0.267.  The  values  of  2h  are,  in  general,  larger  for  the 

H 


f 


>£.tX  A 

;,iv.  V 


*v  r> 


',  •9'- 


■V' 


.•*■/  t>ii 


■\ 


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' 'J 'F’’;^;.'..'  - : ! 

'U  t)i.  '..  '■■ 

s.i:,  f.- : .VI  'Cl  ...:  :>Ls-  i 


f. 

. .''If-.  ' 


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V. 


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' •.  ■•  i , ■ -n.' 

■-.' ■'  ■ -Ji  :'A  r • 


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.......tltl  . . r V*  i*,l 


; ■ . 

i?’’ 

15' 

I 


, I *. 


r <> 


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r .^ 


. r ■ ' . -V  ->  ' a 

.1  I".  ...■%.•  J V 


I ■- 


I ,;  ■ *v 


‘r*. 


‘>jy  ; r ■ '.liYi,: ; $ 


wd-J  ’^a!.  a t;  rf’.lox  ■;  ci&’i. 

♦ * s^:” 

' i,  ‘ 

vvri':.:  a .;• 


iaj^.vr-  I r'-.-  0.1'.:.';  ..'H-C 


•V,  ^ ‘ Ai.  C . . 

d ma-’  ’ r‘*  ‘‘  V r 

*►.4  M ^ ••  Vi  T.  X «u*4  . . < . 


:0. ... . A . J ■.  C* 

ff  ^ ' •* 


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. ,c  a^,  ' •I' 

.-••  ■ too?  >-••.  -ii*; 

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.r  ::4.:  ."u-  \.1‘  z\lQ.zi-iyyi 


r: 


W .'  ' ■ 


-.srrz.'.r. 


52 


fixed  columns  than  for  the  hinged  columns,  and,  in  the  case  of  the 
short  columns,  the  values  of  the  ratio  for  the  fixed  columns  are 
much  larger. 

Due  to  the  great  variations  in  the  values  of  the  ratio  ^ , 
it  is  impossible  to  give  an  average  value  of  the  ratio  that  will 
fit  any  great  number  of  cases.  In  general,  the  values  of  the  ra- 
tio are  smaller  than  the  commonly  assumed  value  0.5,  in  the  case  of 
the  bents  having  longer  columns  and  subject  to  the  smaller  vertical 
loads;  in  the  case  of  the  bents  having  shorter  columns  and  subject 
to  the  larger  vertical  loads,  the  values  of  the  ratio  are  consider- 
ably larger  than  the  value  0.5. 

Points  of  Contraf lexure « 

The  ratio  y/d  is  the  ratio  of  d. i s t an c e f r pm_  b a s e__ p f_  c_o_lumn 

distance  from  base  of  colirmn 

to  point  of  contraf lexure 
to  foot  of  knee  brace 

Graphs  22  to  26  shov/  the  ratios  Y/d  for  both  columns,  due  to 
various  combinations  of  wind  and  vertical  loads,  plotted  with  the 
column  heights  as  abscissas.  The  ratio  Y.^d  for  the  leev/ard  column, 
in  general,  decreases  slightly  as  the  column  height  increases,  due 
to  all  values  of  vertical  load  applied.  There  are  a few  cases, 
hovifever,  where  an  increase  occurs. 

Graphs  27  to  31  give  the  ratios  Y/d  for  both  columns,  due  to 
various  combinations  of  vertical  and  v/ind  loads,  plotted  with  span 
lengths  as  abscissas.  These  graphs  show  that  variations  in  span 
length  produce  no  regular  variations  in  the  ratio  V/d. 

The  location  of  the  point  of  contraf lexure  is  best  studied 

from  Graphs  32  to  36,  which  show  the  ratios  Y/d  for  both  col'amns 

, column  height 

plotted  with  ratios  l~engt h — abscissas. 


53 


The  ratio  for  the  leeward  column  does  not  vary  much  for 

all  values  of  the  ratio  . This  shows  that  the  locatidi 

s^an  xen^xn 

of  the  point  of  contraf lexure  on  the  leeward  column  is  practically 

independent  of  all  characteristics  of  the  bent.  The  lowest  value 

of  is  0.497  for  the  bent  having  the  50-foot  span  and  31- foot 

column  height,  with  a vertical  load  of  100  pounds  per  square  foot. 

The  highest  value  is  0.592  for  the  bent  having  the  20-foot  span  and 

10-foot  column  height,  with  a vertical  load  of  30  pounds  per  square 

foot.  The  average  value  of  the  ratio,  when  there  is  a vertical 

load  of  30  pounds  per  square  foot  (See  Graph  32),  is  0.55.  As  the 

magnitude  of  the  vertical  load  increases,  the  point  of  contra- 

flexure  of  the  leeward  column  is  slightly  lowered.  When  there  is  a 

vertical  load  of  100  pounds  per  square  foot,  the  value  of  the  ratio 

becomes  about  0.51.  These  values  are  very  close  to  the  corrmion- 
d 


ly  assumed  value  of  0.50. 

In  the  case  of  the  windward  column,  certain  combinations  of 
vertical  loads  and  wind  loads  gave  no  point  of  contraf lexure  and 
other  combinations  gave  two  points  of  contraf lexure  (See  Table  XVII), 
In  case  there  were  t\?o  points,  the  point  that  was  nearer  the  average 
value  was  plotted.  Graphs  32  to  36  show  that  there  is  a much  wider 
variation  in  the  values  of  the  ratio  y/d  for  the  windward  column  I 

than  for  the  leeward  column.  This  variation  becomes  greater  as  the 
vertical  load  becomes  larger.  An  average  value  of  the  ratio  ^/d 
for  the  windward  column  is  0.46,  due  to  all  combinations  of  verti- 
cal loads  and  wind  loads  applied. 

15.  Conclusions. 

As  a result  of  this  investigation  the  following  conclusions 


may  be  drawn: 


8£w  sx/Ijav 
it  ni  aott&lrBV 


54 


1.  The  effect  of  the  deformation  of  the  truss  under  vertical 
loads  on  the  reactions  is  too  large  to  he  neglected.  The  values  of 
the  horizontal  reactions  and  moments  due  to  vertical  loads  are  suf- 
ficiently large  to  materially  affect  the  stresses  in  the  knee 
braces  and  truss  members.  The  additional  compressive  stresses  in 
the  knee  braces,  due  to  vertical  loads,  would  necessitate  the  use 
of  a larger  section  for  the  knee  braces.  The  stresses  in  the  mem- 
bers of  the  truss,  due  to  vertical  loads,  are,  in  general,  somewhat 
decreased,  when  the  deformation  of  the  truss  is  considered. 

3,  The  assumption  that  the  horizontal  reactions  are  equal 
is  largely  in  error. 

3.  The  common  assumption^ which  places  the  point  of  contra- 
flexure  midway  between  the  base  of  the  column  and  the  foot  of  the 
knee  brace,  is  not  materially  in  error,  provided,  the  columns  are 
rigidly  fixed  at  their  bases.  In  general,  it  is  slightly  higher 
than  this  mid-point  for  the  lee7,^ard  column  and  slightly  lower  for 
the  windward  col'omn. 

4,  The  resisting  moments,  computed  with  the  assumption  that 
the  horizontal  reactions  are  equal^would  be  materially  in  error,  and 
they  had  best  be  determined  by  an  analysis  of  the  bent. 

The  above  statements  apply  only  to  bents  having  one  quarter 
pitch  Fink  trusses.  Bents  with  other  types  of  trusses  might  now  be 
profitably  investigated. 


55 


VERTICAL  DEFLECTIONS  OF  THE  VARIOUS  PANEL  POINTS 
AND  COL'IPUTATION  OF  THE  REACTIONS  AT  THE  BASE  OF 

EACH  COLUMN 

In  the  following  results,  t,  total  height  of  structure,  = 
height  of  column  + 1/4  of  span  length;  d,  with  a subscript,  in- 
dicates the  vertical  deflection  of  the  panel  point  whose  number  ap- 
or  the  displacement  at  the  base  of  the  right-hand  column 

pears  as  the  subscrip"^’  H,  V^  and  M,  with  a subscript,  but  no  ex- 
ponent, indicate  the  horizontal  reaction,  vertical  reaction,  and 
resisting  moment , respectively , at  the  base  of  the  column  whose  let- 
ter appears  as  the  subscript,  due  to  a uniform  vertical  load  of 
one  pound  per  foot  length  of  truss.  Thus  Hj;^  is  the  horizontal 
reaction  at  the  base  of  the  right  column  due  to  the  uniform  ver- 
tical load.  H,  V,  and  M,  v/ith  both  a subscript  and  an  exponent, 
indicate  the  horizontal  reaction,  vertical  reaction,  and  resist- 
ing moment^  respect ivel3r  at  the  base  of  the  column  whose  letter 
appears  a s the  subscript,  due  to  a single  vertical  load  of  one 
pound  placed  at  the  oanel  point  whose  number  appears  as  the  ex- 
ponent. Thus,  Hj^"^  is  the  horizontal  reaction  at  the  base  of  the 
right  column  due  to  a load  of  one  pound  at  panel  point  Numiber  4. 

d is  the  distance  from  base  of  column  to  foot  of  knee 

brace. 

Y is  the  distance  from  base  of  column  to  point  of  contra- 

flexure  . 

Span  20'  — - Column  height  10' 

Total  vertical  load  = 20  lbs. 

Total  height  of  structure  = t = 10  + 5 = 15  ft. 


I 


Class  I 

Case  A:  d „ r=  6,241 

Z>  ^ 


"3  = 

10 

3^ 

10 

X 

2.5 

= 25 

d = 
0 

615 

0 : 

615 

X 

5 

= 3075 

p. 

CO 

II 

650 

s: 

650 

X 

5 

= 3250 

II 

O' 

615 

Q: 

615 

X 

5 

= 3075 

10 

L: 

10 

X 

2.5 

= 25 

9^450 

^ = 

9450 

- = 1.514 

lbs. 

L 

R 

6241 

«L  - 

1.514 

= 0.1009 

t 

t 

15 

d 


T 


= 755; 

H N 
t t 


755 

6241 


0.13C98 

15 


= 0.12098  lbs. 
0.008065 


56 


Case  3: 


10 

Tr  T 1 X 75 
'^R  ‘ 240 


lb  s . 

~ 

= 0.3125  lbs. 


V 

L 


T 


1 X 165 
240 


V 

0.6875  lbs. 

t 


10  - 
15  “ 

0.5125 

15 

0.6875 

15 


0.6667 

• = 0.02083 

• = 0.04583 


Case  A: 

d.  = 5 
3 

=413 

d„  =405 
s 


Class  II 
dg  = 1972 

j:  5x2.5=  12.5 

o:  413  X 5 = 2065.0 

s;  405  X 5 = 2025.0 


57 


dQ  = 413 
dL  = 5 


t t 
= 465 


H. 


R 


H 


Q:  413  X 5 = 2065.0 

L:  5x2.5=  12.5 


6180.0 


6180 


1372 
3.1339 


= 3.1339  lbs. 


IF 


= 0.2089 


H. 


T = H T 


R 


-iSi-  = 0.23560  lbs. 
1972 


. = 0.01572 

L-  15 


Case  B: 


dp  = 20,374 


= 10  lbs. 

d^=  6,170.3 

y T ^ 6,170.3  _ 

R 20,374 


=_k  = 10  _ C.667 
t t 15 

0.3029  lbs. 


0.3029 

15 


0.02019 


= 1 - 0.3029  = 0.6971  lbs. 


L 

t 

Case  C: 

_ 0.6371 

15 

= 0.04647 

^ 35.55 

= 0.039 

j: 

0 . 039  X 2 . 

5 = 0.10 

<^0 

= 3.12 

o: 

3.13  X 5 

= 15.60 

d 

s 

o 

CO 

• 

03 

II 

s : 

3.80  X 5 

= 14.00 

^0, 

= 2.42 

0: 

2.42  X 5 

= 12.10 

= 0.034 

L: 

0.024  X 2. 

5=  .06 

41.86 


*'1'  \ 


. lU'* 


• * < 


y 


* " 


1% 


r 


t 


« 


59 


Case  A: 


= 11,015 


j: 

12 

X 

2.5 

= 30 

0 : 

800 

X 

5 

= 4,000 

s : 

760 

X 

5 

= 3,800 

Q: 

800 

X 

5 

= 4,000 

L: 

12 

X 

2.5 

= 30 

11,860 

H = H = 1.077  lbs. 
R L 

d = 905 
T 


H = H ^ = 0.0821S  lbs. 
L R 


Case  B; 


='  10  lbs, 


^ 0.3125  lbs. 


\ " = 0.6875  lbs. 


Case  A: 

2 

0 

8 

Q 

L 

Case  3: 


B 

15.0 

2375.0 

2325.0 

2375.0 

8 X 2.5  = 15.0 


Class  II 
d_  = 3,215 


6x2.5 
475  X 5 
465  X 5 
475  X 5 


^ 2.2100  lbs. 


= 532 

= 0.16548  lbs 


*^105.0 


= 10  lbs. 


= 0.3045  lbs. 


dg  = 25,494 

d^  = 7,762.5 


= 0.6355  lbs. 


Cass  C; 


d +A  ^ 39.82 
K 4 


60 


j: 

0.041  X 

2.5  = 

0.10 

0 : 

3.18  X 

5 

15.90 

s : 

2.72  X 

5 

13.60 

Q,: 

2.51  X 

5 

12.55 

L: 

0.027  X 

2.5  = 

0.07 

42.22 

M = M-  = -1-^-^-—  = 101.79  in.  lbs. 

^ L 39.82 

= 3.58 

= 8.631  in.  lbs.  = 6.711  in.  lbs. 

Yp^  = Y,  = ..  lp,l «-7-g-  :=  46.06  in. 

^ 2.21 

Y = 52.16  in.  Y,  = 40.55  in. 

R L 


Span  20'  Column  Height  16' 

t = 21« 

Class  I 

Cass  A;  = 27,387 


15 

X 

2.5 

LO 

• 

to 

II 

= 0.567  lbs. 

0 : 

1050 

X 

5 

= 5,250.0 

= 1,182 

s: 

990 

X 

5 

= 4,950.0 

T 

Q: 

1050 

X 

5 

= 5,250.0 

= 0.04316  lbs. 

L: 

Case  B: 

15 

X 

2.5 

= 37.5 

15,525.0 

= 10  lbs. 

= 0.3125  lbs.  = 0.6875  lbs. 


f 


\ 


/ 


H 


H 


I 

i 

II 


I 


I 


A 

t 


: • V 

M 


I 


\ 


61 


Class  II 


A; 

S = 

7,410 

j: 

8 

X 

2.5 

= 20.0 

It 

0: 

595 

X 

5 

= 2975.0 

dm  = 

8 : 

580 

X 

5 

= 2900.0 

T 

Q: 

595 

X 

5 

= 2975.0 

H T 

L: 

8 

X 

2.6 

= 20.0 
8,890.0 

R 

= H = 0.09042  lbs. 
L 


B: 

d = 35,734 
B 

V ^ 
R 

= = 10  lbs. 

d^  = 10,962.5 

= 0.3068  lbs. 

V = 0.6932  lbs. 
L 

C: 

d^  = 78.133 

2' 

0.041  X 2.5  = 

0.10 

0: 

3.18  X 5 = 

15.90 

s: 

2.82  X 5 = 

14.10 

Q: 

2 . 62  X 5 — 

13.10 

L: 

0.029  X 2.5  = 

0.07 

43.27 


. M . 43^27„x  14£_  ^ ,0.75  in.  lbs. 

78.133 


R "L 

dip  s 3 . 56 

T 


= 6.561  in.  lbs. 

Y = = ■J!9..7L-  = 66.47  in. 

R L 1.1S97 


M/  = 5.195  in.  lbs 

Ij 


Y^^  = 72.56  in. 


Yj^^  = 57.43  in. 


62 


Span  20’  Column  Height  19' 

t = 24  feet 
Class  I 


Case  A 

J 

0 

S 

Q 

L 


= 71, 

18  X 2.5  = 45 

1170  X 5 = 5850 

1160  X 5 = 5800 

1170  X 5 = 5850 

18  X 2.5  = 45 

17,590 


419 

= 0.246  lbs. 

= 1,380 

H = H 1 r:  0.01932  lbs. 
R L 


Case  B: 


= Vl  = 10  lbs. 

V = 0.3125  lbs.  = 0.6875  lbs. 

^ L 


Case  A: 

J:  9 X 2.5 

0:  695  X 5 

S:  665  X 5 
Q. : 695  X 5 
L:  9x3.5 

Case  B: 


Class  II 
dg  = 13,450 

33.5  Hp^  = \ = 0.829  lbs. 

3475.0 

J 

3325.0  T = 790 

3475.0 

H T = H T = 0.05345  lbs. 

32.5  ^ ^ 

10.320.0 

d = 43,414 

•D 


\ ^ \ lbs,  d = 13,362.5 

1 

V 1 = 0.3078  lbs.  V I"  = 0.6922  lbs. 

R L 


• 5 . * w 

A 


!B 


63 


C: 

d 

K 

+ 

It 

J: 

0.042 

X 2 . 5 = 

0.11 

0: 

3.09 

X 5 = 

15.45 

S: 

2.65 

X 5 = 

13.25 

Q: 

2.60 

X 5 = 

13.00 

L: 

0.029 

X 2. 5 = 

.07 

41.88 

IvL 

= M_  = 

X 

00 

00 

1 — 1 

180  _ ; 

115.687 


diji  — 3 . 43 


= 5.337  in.  lbs.  M T ==  4.209  in.  lbs. 
^ L 


Y_  ^ V - 65.16 

“ - 78. oO  in. 


0,829 


= 84.11 


^ 66.34  in. 


S4 


Span  30'  Column  Height  16' 

Total  vertical  load  on  truss  = 30  lbs. 
t = 23.5  feet. 

Class  I 

Case  A:  d_  = 45,914 

3 


J: 

23 

X 

1.875 

= 

43.13 

II 

Hj.  = 1.062  lbs. 

2: 

1440 

X 

3.75 

5,400.00 

"^4  = 

2,035 

3: 

1925 

X 

3.75 

= 

7,218.75 

6; 

2160 

X 

3.75 

= 

8,100.00 

H 4 
R 

= H ^ ==  0.04432  lbs. 
L 

7: 

1925 

X 

3.75 

= 

7,218.75 

^1  = 

1,650 

10: 

2160 

X 

3.75 

s 

8,100.00 

= H.  ^ = 0.03594  lbs. 

11: 

1925 

X 

3.75 

= 

7,218.75 

R 

L 

13: 

1440 

X 

3.75 

= 

5,400.00 

L: 

23 

X 

1.875 

J2; 

43.13 

48,742. 51 


Case  B: 

= 15  lbs. 


V 4 _ lx  113.5 
R 360 


0.3125  lbs. 


1 X 247.5 
360 


1 X 58.35 
360 


1 X 305.75 


0.6875  lbs. 
0.15625  lbs. 

: 0.84375  lbs. 


360 


JlIV 


Case  A: 


Class  II 


d_  = 12,530 


65 


J: 

12 

X 

1.875 

= 22.50 

II 

2; 

830 

X 

3.75 

= 3,112.50 

d4  = 

3: 

1085 

X 

3.75 

= 4,068.75 

6: 

1230 

X 

3.75 

= 4,612.50 

7: 

1100 

X 

3.75 

= 4,125.00 

10: 

1230 

X 

3.75 

= 4,612.50 

H 1 

11: 

1085 

X 

3.75 

= 4,068.75 

R 

13: 

830 

X 

3.75 

= 3,112.50 

L: 

13 

X 

1.875 

= 22.50 

27,757.50 

i B: 

dg  = 133,714 

,07662  lbs. 


R 


Case  C: 


= 15  lbs. 

u L 

= 41,243.9 

0.3084  lbs. 


V. 


4 _ 


= 0.6916  lbs. 


d^  = 20,195.9 


V ■*•  = 0.1510  lbs. 
R 


= 0.8490  lbs. 


+ = 130.325 


J: 

0.072 

X 

1.875 

= 

0.135 

2: 

4.60 

X 

3.75 

= 

17.250 

3: 

5.  75 

X 

3.75 

= 

21.563 

6: 

6.27 

X 

3,75 

= 

23.512 

7: 

5.10 

X 

3.75 

- 

19.125 

10: 

5.28 

X 

3.75 

19.800 

• • 

f — 1 
rH 

4.39 

X 

3.75 

= 

16.463 

13: 

3.08 

X 

3.75 

= 

11.550 

L: 

0.03? 

X 

1.875 

mZ 

0.069 

129.467 


66 


^ 139 » 467  X 144 


\ 130.225  = 143.15  in.  lbs. 


^4  = 6 

>.09 

<^1  = 

5.33 

1 

in. 

Mr4  = 

6.734 

in.  lbs. 

» 

II 

< 

CD 

00 

in 

Ibi 

II 

5.258 

in.  lbs. 

M 1 = 

L 

4.004 

in 

a Y = 

. 143.16 

= 64.62 

in. 

R 

2.2153 

T 

II 

>* 

73.37 

in. 

76. 

93 

in 

57.29 

in. 

V = 

52. 

26 

in 

Span  30' 

-Col.  Height 

21' 

t = 28.5  ft. 
Class  I 


Case  A: 


d_  = 108^344 
5 


J 

30 

X 

1.875  = 

56.25 

2 

1840 

X 

3.75  = 

6,900.00 

R 

3 

2380 

X 

3.75  = 

8,935.00 

6 

3670 

X 

3.75  = 

10,012.50 

7 

2320 

X 

3.75  = 

8,700.00 

10 

2670 

X 

3.75  = 

10,012.50 

R 

11 

2380 

X 

3.75  = 

8,925.00 

d,  = 

13 

1840 

X 

3.75  = 

6,900.00 

1 

L 

30 

X 

1.875  = 

56.25 

= 

60,487.50 

R 

4 _ 


Case  B: 


V = V = 15  lbs. 

it  L 

= 0.3125  lbs. 
= 0.6875  lbs. 


V_^  ==  0.15625  lbs. 
R 

V_^  = 0.84375  lbs. 
L 


5583  lbs. 


0.02358  lbs. 


0.01957  lbs. 


r 


» 


\ 


/ 


Class  II 


67 


Case  A:  dg  = 28^372 


J 

18 

X 

1.875 

=S 

33.75 

2 

1070 

X 

3.75 

= 

4012.50 

3 

1370 

X 

3.75 

5137.50 

6 

1500 

X 

3.75 

s 

5625.00 

7 

1300 

X 

3.75 

4875.00 

10 

1500 

X 

3.75 

5625.00 

11 

1370 

X 

3.75 

=S 

5137.50 

13 

1070 

X 

3.75 

4012.50 

L 

18 

X 

1.875 

33.75 

34,492.50 

= Ht  = 1.2157  lbs. 

= 1420 

= 0.05005  lbs. 

d^  = 1200 

Hi  = H 1 = 0.04230  lbs. 
R L 


Case  B: 


(ig  = 181,595 


^R  = 

V = 15  lbs. 
L 

II 

56,206.3 

d^  = 27,677.; 

I 

V 

= 0.3095  lbs. 

= 0.1524 

lbs. 

=*  0.6905  lbs. 

= 0.84  76  lbs. 

Case  C: 

^ +^4 

= 241.020 

J: 

0.071  X 1.875  = 

0.133 

2: 

4.45  X 3.75  = 

16.688 

3: 

5.55  X 3.75  = 

20.812 

6: 

6.08  X 3.75  = 

22.800 

7: 

4.92  X 3.75  = 

18.450 

10: 

5.00  X 3.75  = 

18.750 

11: 

4.30  X 3.75  = 

16.125 

13: 

3.05  X 3.75  = 

11.438 

L: 

0.038  X 1.875  = 

0.071 

125.267 

Mt  = 

„ - 125.267  X 

204  = 106.03 

in.  lbs. 

L 

241.020 

II 

5.93 

= 5.18 

V 

- 5.019  in.  lbs. 

^ 4.384 

in.  lbs. 

3.939  in.  lbs. 

Mgl  = 2.998 

in.  lbs. 

1. 

R H - 1.2157 

- = 87.22  in. 

Y 4 
R 

= 100.28  in. 

= 103.64 

in. 

Y 4 
L 

==  78.70  in. 

==  70.87 
L 

in. 

I 


V- 


• .i 


V 


'/ ' 


Y*' 

.A 


i 

i 


68 

Span  30'  Column 

Height  26' 

t = 33.5  ft. 

Class  I 

Case  A: 

dg  = 212,139 

J 

37  X 1.875  = 69.38 

Hp  = H^:=  0.3560  lbs. 

2 

2250  X 3.75  = 8,437.50 

3 

6 

3000  X 3.75  = 11,250.00 
3325  X 3.75  = 12,468.75 

d^  = 3150 

7 

2950  X 3.75  = 11,062.50 

10 

3325  X 3.75  = 12,468.75 

K^4  » N 4 ^ 0.01485  lbs. 

11 

3000  X 3.75  = 11,250.00 

irt  L 

13 

2250  X 3.75  = 8,437.50 

d^  = 2560 

L 

37  X 1.875:=  69,38 

1 

75,513.76 

= H ^ = 0.01207  lbs. 

R L 

Case  B: 

= 15  lbs. 
= 0.3125  lbs. 

= 0.15625  lbs. 

V_^  = 0.6875  lbs. 

i)  Jj 

= 0.84375  lbs. 

Class  II 

Case  A 

d^  = 54,557 

J 

19  X 1.875  = 35.63 

= H,  = 0.7852  lbs. 

R 

2 

1280  X 3.75  = 4800.00 

3 

1640  X 3.75  = 6150.00 

d.  = 1720 

6 

1830  X 3.75  = 6862.50 

4 

7 

10 

1590  X 3.75  = 5962.50 

1830  X 3,75  = 6862.50 

= 0.03153  lbs. 

11 

1640  X 3.75  = 6150.00 

d^_  = 1460 

13 

1595  X 3.75  = 5981.25 

L 

19  X 1.875  = 35.63 

= 0.02676  lbs. 

42,840.01 

R L 

Case  B 

: dg  = 229,477 

= 15  lbs. 

d, 

^ = 71,168.9  d^_  = 

35,158.4 

^ 0.3101  lbs. 

= 0.1532  lbs. 

= 0.6899  lbs. 

h Jj 

= 0.8488  lbs. 

t 


f ♦ 


r 


'.T 


If 


■i 

I . 


• X > 

y '■  . ^ 


i ’ 


/.  w % ^ 

X CC 

X Ov- 

■r,  V<  M 


/. 


I 

1 


I 


j. 


.jf 


L. 


■ I 


t 


J 


/ 


^ t 

I 


I 


4 


; ^ 


'uX 


\ 


. '.  . . I'f.  .-i* 


/J3 


;..  e-a^r. 


a . 

ri^ ' 

f ..  : ., 

)i 


i 


. - -f  l 


■aaeaEeCT?.- 


Case  C:  + ■A4  = 383.494 


J: 

0.070 

X 

1.875  = 

0,131 

2: 

4.30 

X 

3.75  = 

16.125 

3: 

5.35 

X 

3.75  = 

20.063 

6: 

5.65 

X 

3.75  = 

21.187 

7: 

4.40 

X 

3.75  = 

16.500 

10: 

4.58 

X 

3.75  •= 

17.175 

11: 

3.98 

X 

3.75  = 

14.925 

13: 

2.75 

X 

3.75  = 

10.313 

L: 

0.037 

X 

1.875  = 

0.069 

116.488 

= M.  = 

264  X 116. 

= 79.16  in.  lbs. 

R 

L 

388.494 

d4  = 

= 5.58 

d^  = 4.82 

V 

= 3.792 

in.  lbs. 

= 3.275 

in. 

lbs 

= 2.928 

in.  lbs. 

IL^  2.177 

in. 

lbs 

= Yt  = _Z£il6  ^ 100.82  in. 

0.7852 

Y„^  = 130.27  in.  Y ^ 122.38  in. 

R R 

Yj^^  = 92.86  in.  Y^^^  = 81.35  in. 

Span  30’  — - Column  Height  31' 
t = 38.5  ft. 

Class  I 

Case  A:  d_  = 164,493 

3 


J: 

44 

X 

1.875 

82.50 

II 

0.5322  lbs. 

2: 

2645 

X 

3.75 

= 

9,918.75 

Jj 

3: 

3480 

X 

3.75 

=: 

13,050,00 

d^  = 

3700 

6: 

3875 

X 

3.75 

= 

14,531.25 

4 

7: 

3300 

X 

3.75 

35; 

12,375.00 

tr  4 

TT  4 

= 0.02249  lbs 

10: 

3875 

X 

3.75 

s 

14,531.25 

% 

= 

11: 

3480 

X 

3.75 

= 

13,050.00 

dn  — 

3025 

13: 

2645 

X 

3.75 

=£ 

• 9,918.75 

^1 

L: 

44 

X 

1.875 

= 

82.50 

= H 1 

= 0.01839  lbs 

87,540.00 


70 


Case  B: 


^ = 15  lbs. 

Vr^ 

= 0.15625  lbs 

V 

= 0.3125  lbs. 

= 0.84375  lbs 

= 0.6875  lbs, 

Class  II 

Case  A: 

= 42,340 

J: 

18  X 1.875  = 

33.75 

Hr  = % = 1. 

2: 

1370  X 3.75  == 

5,137.50 

3: 

1800  X 3.75  = 

6,750.00 

d.  = 1900 

6: 

2000  X 3.75  = 

7,500.00 

fr 

7: 

10: 

1760  X 3.75  = 

2000  X 3.75  = 

6,600.00 

7,500.00 

Hr^  = = 

11: 

1800  X 3.75  = 

6,750.00 

= 1580 

13: 

1370  X 3.75  = 

5,137.50 

L: 

18  X 1.875  = 

33.75 

_ = H 1 = 

R L 

45,442.50 

Case  B: 

S = 

= 105,261 

II 

= V_  =15  lbs. 
L 

p* 

II 

= 32,354.1 

di  = 

= 15,751.0 

= 0.3074  lbs. 

V 1 
R 

= 0.1496  lbs. 

= 0.6926  lbs. 

= 0.8504  lbs. 

Case  C:  ^4  ""  228.323 


J: 

0.047 

X 

1.875 

r= 

0.088 

2: 

3.57 

X 

3.75 

sz. 

13.388 

3: 

4.36 

X 

3.75 

16.350 

6: 

4.65 

X 

3.75 

=: 

17.437 

7: 

3.  73 

X 

3.75 

s 

13.988 

10: 

3.77 

X 

3.75 

sr 

14.137 

11: 

3.20 

X 

3,75 

S£ 

12.000 

13: 

2.37 

X 

3.75 

s= 

8.888 

L: 

0.013 

X 

1.875 

0.024 

96.300 


= g34_x._9J.._5p_Q  ^ 140.03  dn.  lbs. 
R L 222.822 


0733  lbs. 


0.04487  lbs. 


0.03732  lbs. 


71 


= 4.51 


= 4.06 


= 6.558  in.  lbs. 
H 


4.722  in. 


lbs. 


V = 

M 1 r= 
L 


= 140.05  ^ 

1.0733 

\ 

= 146.16  in. 

Y/^  = 105.24  in. 

L 


130.47  in. 


L 


5.904  in. 
3.510  in. 


158.20  in 
94.05  in 


lbs. 

lbs. 


72 


Case  A: 


Span  40*  Column  Height  16* 

Total  vertical  load  on  truss  = 40  lbs. 
t = 26  feet. 

Class  I 
dg  = 32,309 

= 1.897  lbs. 


J:  25  X 2. 5 = 62,5 

2:  1410  X 5 = 7,050.0 

3:  1840  X 5 = 9,200.0 

6:  2025  x 5 =10,125.0 

7:  1800  X 5 = 9,000.0 

10:  2025  X 5 =10,125.0 

11:-  1840  X 5 = 9,300.0 

13:  1410  X 5 = 7,050.0 

L:  25x2.5=  62,5 


61,875.0 


d4  = 1970 

= 0.06041  lbs. 

d^  = 1590 

H ^ = H ^ = 0.04876  lbs. 
R L 


Case  B: 


V. 


R 


V_  = = 20  lbs, 

K L 

= 0.3125  lbs.  vj-  = 1 X 75  ^ 

R 


Case  A: 


40 

= l-X,  27.. 5 0.6875  lbs. 

40 

Class  II 
dg  = 9,445 


= 0.15625  lbs. 


480 


V, 


1 = 1 X 405  , 
' 480 


0.84375  lbs 


J: 

15 

X 

2.5 

=: 

37.5 

2: 

910 

X 

5 

4550.0 

3: 

1190 

X 

5 

= 

5950.0 

6: 

1320 

X 

5 

s: 

6600.0 

7: 

1150 

X 

5 

ss 

5750.0 

10: 

1320 

X 

5 

= 

6600.0 

11: 

1190 

X 

5 

5950.0 

13: 

910 

X 

5 

= 

4550.0 

L: 

15 

X 

2.5 

= 

37.5 

4C 

>,025.0 

d4  = 1260 

= 0.13340  lbs. 

= 1040 

H 1 = H ^ = 0.11012  lbs. 
R L 


Case  B: 
V. 


dg  = 211,776 


R 


= = 30  lbs. 


d^  = 65,530,5 


d^  = 32,169.4 


V. 


R 


= 0.3094  lbs. 


V. 


R 


= 0.1519  lbs. 


V,  = 0.6906  lbs. 


V 1 = 0.8481  lbs. 
L 


73 


Case  C: 

. + 

A _ 
4 “ 

101. 

27 

J; 

0.098 

X 

2.5 

= 0 

.245 

2: 

5.55 

X 

5 

= 27 

.750 

3: 

6.95 

X 

5 

= 34 

.750 

6: 

7.48 

X 

5 

= 37 

.400 

7: 

6.20 

X 

5 

31 

.000 

10: 

6.28 

X 

5 

= 31 

.400 

11: 

5.50 

X 

5 

= 27 

.500 

13: 

3.78  ; 

X 

5 

= 18 

.900 

L: 

0.056 

X 

2.5 

= 0 

.140 

209 

.085 

\ - 

= M = 
L 

209.085  X 
101.27 

120 

= 247. 

.76  in. 

lbs. 

- 

= 7.35 

= ( 

5.24 

V 

= 8.709 

in. 

lbs. 

II 

7.394 

in,  lbs. 

= 7.221 

in. 

lbs. 

M^l  = 

5.306 

in.  lbs. 

= Y = = 58.47  in. 

^ ^ 4.3377 


Yj^^  = 65.28  in. 


Y_^  = 67.14  in. 
K 


Y^^  = 54.13  in. 


Y = 48.18  in. 
L 


Span  40*  Column  Height  21’ 

t = 31  feet. 

Class  I 


Case  A: 


d = 83,483 

jD 


J 

30 

X 

2.5 

= 

75 

2 

1790 

X 

5 

8,350 

3 

2390 

X 

5 

= 

11,950 

6 

2640 

X 

5 

= 

13,200 

7 

2240 

X 

5 

r= 

11,200 

10 

2640 

X 

5 

= 

13,200 

11 

2390 

X 

5 

= 

11,950 

13 

1790 

X 

5 

r: 

8,950 

L 

30 

X 

2.5 

=: 

75 

79,550 


0.9523  Ihs. 

= 2520 

= 0.03018  lbs. 

d^  = 2000 

= 0.02396  lbs. 


r« 


74 


Case  B: 


= 20  lbs. 


V_^  = 0.3125  lbs. 
K 


= 0.6S75  lbs. 


= 0.15625  lbs. 


V ^ = 0.84375  lbs. 
L 


Case  A: 


Class  II 
A = 22,346 

D 


J: 

19 

X 

2.5 

S= 

47.5 

2; 

1070 

X 

5 

= 

5350.0 

3; 

1400 

X 

5 

7000.0 

6: 

1570 

X 

5 

7850.0 

7: 

1380 

X 

5 

=: 

6900.0 

10: 

1570 

X 

5 

7850.0 

11: 

1400 

X 

5 

=: 

7000.0 

13: 

1070 

X 

5 

5350.0 

L: 

19 

X 

2.5 

47.5 

H 


R 


= = 0.06632  lbs. 


47,395.0 


= 1240 

H ^ = Hi  = 0.05549  lbs. 
K L 


Case  B: 

Vj^  = = 20  lbs. 

d4  = 92,132.7 


d = 296,899 
B 


R 


= 0.3103  lbs. 


= 0.6897  lbs. 
L 


d^  *=  45,470.5 

V_^  = 0.1532  lbs. 

K 

= 0.8468  lbs. 


Case  C: 


dx^  +A^  = 198.99 


J: 

0.091 

X 

2.5 

==  0.228 

2: 

5.05 

X 

5 

= 25.250 

3: 

6.55 

X 

5 

= 32.750 

6: 

7.00 

X 

5 

= 35.000 

7: 

5.85 

X 

5 

= 29.250 

10: 

6.12 

X 

5 

= 30.600 

11: 

5.45 

X 

5 

= 27.250 

13: 

4.21 

X 

5 

= 21.050 

L: 

0.051 

X 

2.5 

= 0.127 

201.505 

“r 

= M = 

180  X 

201.505  . 

L 

198.99 

<^4 

= 3.85 

1 

= 5.78 


I 


aagjjjjjJ 


or,  . 


r 


I 


•>c  •*  ^ 


4- 


\ 


/. 


r 

I 


1 


•j  ^ ■ ..I 


i 


^ • / *> 


* * •. 


I V- 

* • • 4 


. 7. 

':  r 

3 i' 

c i 


- A . 


1 


u 


■ . 


' ' ^ 


•'i 

UftJ 


A 


najiTi  rjinr  I ~ii''~ 


75 


II 

6. 

196 

in. 

lbs, 

V 

= 5.228 

in. 

lbs 

II 

5. 

140 

in. 

lbs. 

= 3.764 

in. 

lbs 

Y 

“T) 

182.37  _ 

85.94  in. 

n 

L 

2.121 

II 

93 

.43 

in> 

V 

= 94.22 

in. 

v = 

77 

.50 

in. 

= 67.83 

in. 

-76- 


Span  40'  - Col.  Height  26' 
t = 06  ft. 

Class  I. 


Case  A: 


cl  = 172,255 
B 


J; 

30 

X 

2 1/2 

75 

2: 

2250 

X 

5 

= 

11,250 

3: 

2925 

X 

5 

= 

14,625 

6: 

3350 

X 

5 

= 

16,250 

7: 

2825 

X 

5 

= 

14,125 

10: 

3250 

X 

5 

= 

16,250 

11: 

2925 

X 

5 

= 

14,625 

13: 

2250 

X 

5 

sr 

11,250 

L: 

30 

X 

3.1/3 

zz 

75 

98,525 

Case  B; 
V 

R 

r4 


V’  = 20  lbs. 
L 


= 0.3125  lbs. 
R 

Vr^  = 0.6875  lbs. 


H = H = 0.5720  lbs. 

R L 

d = 3,100 

= 0.01600  lbs. 

d.  = 3,575 
i T-1 

K = H = 0.014S5  lbs. 
R L 


V = 0.15625  lbs. 
R 

= 0.84375  lbs. 
L 


Class  II. 


Case  A 


d = 44,707 
B 


JJ 

23 

X 

2. 

5 = 57.5 

2: 

1320 

X 

5 

=6,600.0 

3: 

1750 

X 

5 

= 8,750.0 

6: 

1910 

X 

5 

= 9,550.0 

7: 

1700 

X 

5 

= 8,500.0 

10: 

1910 

X 

5 

= 9,550.0 

11: 

1750 

X 

5 

= 8,750.0 

13: 

1320 

X 

5 

= 6,600.0 

L : 

23 

x2.5 

= 57.5 

58,415.0 


= H,  = 1 . 3066  lbs . 


R 


d,  = 


= h;  = 

R L 

d = 
1 

1 1 

H = H = 
R L 


i,S20 

0.04071  lbs. 
3,500 

0.03355  lbs. 


" ■ I .■(  i‘‘  I 

> ' . ■ ■ -V  r 


.1 


-77 


Case  B: 


= V 
d 

V 

V 


d = 382,02^ 
B 

= 20  lbs. 

= 118,732.7 

= 0.3108  lbs. 
= 0.6892  lbs. 


d^  = 58,770.5 

= 0.1538  lbs. 
R 

L = 0.6462  lbs. 


Case  C:  + ^4  = 332.11 


J: 

O.OSO 

X 

2.5 

= 

0.23 

2 

4.90 

X 

5 

= 

24.50 

3 

6.22 

X 

5 

= 

31.10 

r> 

D 

6 . 60 

X 

5 

= 

33.00 

7 

5 . -10 

X 

5 

= 

27.00 

10 

5.56 

X 

5 

27.75 

11 

4.85 

X 

5 

24,25 

13 

3.45 

X 

5 

= 

17.25 

L 

0.052 

X 

2.5 

= 0.13 

185.21 

R 

Ti/r-i 


M = M = 240  X 185, 

<=j1  = 84  in  Ih R . 

R L S32.il 

= 6.35 

dj 

=5.62 

= 4.589  in  lbs. 

Mr 

= 4.061  in  lbs. 

= 3.773  in. lbs. 

L 

= 2.885  in  lbs 

Y = Y = 133.84  = 

R L T73U5^ 

102.43 

in. 

= 113.72  in. 

1 

Y 

R 

= 121.04  in. 

= 92.68  in. 

= 85,99  in. 

L 


Case 


Case 


Case 


Span  40'  - Col.  He 
t = 41  ft. 
Class  I. 

A:  d = 135 , 181 

B 


J:- 

40 

X 

2 

1/2 

100 

2: 

2600 

X 

5 

= 

15,900 

3450 

X 

5 

= 

17,250 

6: 

3825 

X 

5 

= 

19,125 

7: 

3350 

X 

5 

= 

16 , 750 

10: 

3825 

X 

5 

= 

19,125 

11: 

3450 

X 

5 

= 

17,250 

13: 

2600 

X 

5 

= 

13,000 

L: 

40 

X 

2 

1/3 

= 

100 

115,700 


B: 

V = V =20  lbs. 

R L 

= 0.3125  I'os. 

= 0.6875  lbs, 

L 

Class  II. 


-78 

ght  31' 


H = H = 0.8559 
R L 

d = 3610 

H^=  = 0.02670  Ids. 

R L 

d;j_=  3000 

= 0.02219  lbs. 

R L 


= 0. 15325  lbs. 
R 

= 0.84375  lbs. 
L 


A:  d = 35,685 

B 


21 

X 

2.5 

= 52.5 

2: 

1425 

X 

5 

= 7,125.0 

o : 

1890 

X 

5 

= 9,450.0 

6: 

2100 

X 

5 

=10,500.0 

7: 

1890 

X 

5 

= 9,450.0 

10: 

2100 

X 

5 

=10,500.0 

11: 

1890 

X 

5 

= 9,450.0 

13: 

1425 

X 

5 

= 7,125.0 

L: 

21 

X 

2.5 

= 52.5 

63,705.0 


H = H = 1.7852  lbs. 

R L 

d = 2010 

4 

= 0.05633  lbs. 

R L 

d^  ^ 1650 

1 1 

K = = 0.04624  lbs. 


-79- 


Case  B: 

d = 175,594 
B 

>' 

II 

> 

20  lbs. 

R L 

d 

= 54,228.3 

d = 26,518.3 

4 

1 

R 

= 0 . 3088  lbs. 

= 0.1510  lbs 
'^R 

1 

= 0,6912  lbs. 

V 

L 

L = 0.8490  lbs 

Case  C: 

d. 

+ A = 196.29 

K 4 

J :*• 

0 . 063 

X 2 . 5 = 0.16 

2: 

3 . 98 

X 5 = 19.90 

3 : 

5.00 

X 5 =25.00 

6: 

5 . oO 

X 5 = 26..50 

7: 

4.27 

X 5 =21.35 

10: 

4. 45 

X 5 =22.25 

11: 

3.75 

X 5 =18.75 

13: 

2 . 

X 5 = 12.75 

L: 

0.030 

x2 « 5 = , C 8 

r4S . TT' 

1 

= M 

= 146.74  X 300  = 

224.27  in.  lbs. 

R L 

196 . 29 

d 

4 

= 5 , 22 

d^  = 4.52 

= 7.978  in  lbs. 

= 6.202  in.  lbs. 
L 


'aI  = 6.908  in,  lbs. 
R 

1 

M = 4.5S8  in.  lbs. 
L 


Y = Y = 224.37 
^ ^ 1.7852 

Y""  = 141.6b  in. 

R 


125. So  in. 


1 

Y = 149.39  in. 
R 


Y^  = 110.10  in. 
L 


Y = 94.90  in. 
L 


-80- 


Span  50'  - Col.  Height  16  * 

Total  Vertical  Load  on  Truss  = 50  lbs. 


t = 28.5  ft. 


Class  I. 


Case  A: 


= 29,654 


J« 

• 20 

X 

3. 125 

= 62.50 

2 

1170 

X 

6.25 

= 7,312.50 

3 

15  30 

X 

6.25 

= 9,562.50 

6 

1695 

X 

6.25 

10, 593. 75 

7 

1455 

X 

6.25 

= 9,093.75 

10 

1695 

X 

6.25 

= 10,593.75 

11 

15  30 

X 

6.25 

= 9,562.50 

13 

1170 

X 

6.25 

= 7,312.50 

L; 

20 

X 

3. 125 

= 62.50 

64,156.  25 


H 

R 


H 


I 

R 


H = 2.1635  lbs. 

L 

, = 1641 

= 0.055  34  lbs. 
L 

d = 1344 

= 0.045  32  lbs. 
L 


Case  B: 


V = V =25  lbs. 

R L 

= 1 X 187.5  = 0.3125  lbs. 
^ 600 


= 1 X 93.75  = 0.15625  lbs 

= 1 X 506.25  = 0.84375  lbs 
^ 600 


Case  A: 


JL-X  A^S.5  ^ 0.6875  lbs. 

600 

Class  II 

=8,369 


H = H = 5.1694  lbs. 


12 

X 

3. 125 

=r 

37.50 

R 

L 

2 

760 

X 

6.25 

4,750.00 

3 

1045 

X 

6.25 

=s 

6,531.25 

d =1120 

6 

1130 

X 

6.25 

= 

7,062.50 

4 

7 

1040 

X 

6.25 

:s 

6,500.00 

10 

1130 

X 

6.25 

= 

7,062.50 

= 0.1338  3 

OQ 

1 — 1 

11 

1045 

X 

6.25 

=s 

6,531.25 

R 

L 

13 

760 

X 

6.25 

4,750.00 

d = 880 

L- 

12 

X 

3. 125 

— 

37.50 

1 

ll 

4 3, '262. 50 

H = 

H = 0.10515 

lbs 

R 

L 

-31- 


Case  B: 


d = 327,902 
B 


V V =25  lbs. 
R L 


d = 101,704.7 

4 


v^-  = 

R 

4 

V = 


0.3102  lbs. 
0.6898  lbs. 


V, 


R 

1 


50,292.7 

0.1534  lbs. 
0.8466  lbs. 


Case  C: 


d + ^ 
K 


= 97..47 


J 

0.078 

X 

3.125 

= 0.24 

2 

5. 10 

X 

6.25 

= 31.88 

3 

6.  60 

X 

6.25 

= 41.25 

6 

7. 10 

X 

6.25 

=*  44.37 

7 

6.03 

X 

6.25 

= 37.69 

10 

6.  33 

X 

6.25 

= 33.56 

11 

5.50 

X 

6.25 

= 34.38 

13 

3.95 

X 

6.25 

= 24.69 

L 

0.045 

X 

3.  125 

= 0.14 

254.20 

M 

= M = 

254. 20 

X 120  = 

= 312.96  in.  lbs. 


R 


97,47 


d = 

6.90 

= 5.90 

4 

8,495 

in. 

lbs. 

1 

M = 7.264 

in. 

lbs 

R 

R 

= 

7.115 

in. 

lbs. 

= 5.554 

in. 

IbB 

L 

L 

Y = 

Y 

= 312.96 

= 60.54  in. 

R 

L 

5 . 1694 

A. 

1 

Y = 

63.48 

in. 

Y = 63.08 

in. 

R 

R 

=* 

5 3. 16 

in. 

Y^  = 52.82 

in. 

L 

L 

-lB- 


( 


* ' r'lrs  ■ - '■ 


/• 


i 


' ^ ■ '■ 


' 

• 

A • 

*t, 

.•  . 

/'. 

- , 

, ' 

X 

. 

> 

et’ 

'V  £ 

■i 

fix-  . 

- ,•■' . 

- 

c^  V 

, 

. . . ! 

. ^ 

X 

■ - 

X 

■| 

. ‘ . ♦ r • ' • - ^ 

' »iVi'  ' ' *•  f ' ~ 


<r 

? aw 


I 


~-y--‘gr 


-82- 


Case 


Case 


Case 


Span  50'  - Col.  Height  21* 
t = 33.5  ft. 

Class  I. 


A: 

78,455 

J 

27 

X 

3.  125 

= 84.38 

H.  = 1,0370  lbs. 

2 

1480 

X 

6.25 

= 9,250.00 

L 

3 

1960 

X 

6.25 

= 12,250.00 

= 2,0bb 

6 

2120 

X 

6.25 

= 13,  250.00 

4 

7 

1870 

X 

6.25 

=11,687.50 

h"  = 

= 0.02619  lbs 

10 

2120 

X 

6.25 

=13,250.00 

R 

L 

11 

1960 

X 

6.25 

= 12,250.00 

13 

1480 

X 

6.25 

= 9,250.00 

dj  = 1720 

L 

27 

X 

3. 125 

.84_.  5S 

= 

H = 0.02193  lbs 

81,356;26 

R 

L 

B: 

V 

= V = 

25  lbs 

• 

R L 

1 

^ = 0. 3125 

lbs. 

= 0.35625  lbs. 

R 

R 

1 

4 

V 

" = 0.6875 

lbs. 

Vl 

= 0.  84375  lbs. 

Class  II. 


A: 

20,639 

J 

16 

X 

3. 125 

50.0 

H = H = 

2.5177 

lbs. 

2 

940 

X 

6.25 

5,875.0 

R L 

3 

1240 

X 

6.25 

— 

7,750.0 

3,300 

6 

1364 

X 

6.25 

8,525.0 

d = 

7 

1210 

X 

6.25 

— 

7,562.5 

Ar 

10 

1364 

X 

6.  25 

=: 

8,525.0 

«R  = = 

0.06299 

lbs. 

11 

1240 

X 

6.25 

7,750.0 

13 

940 

X 

6.25 

=: 

5,875.0 

d,  = 

3,080 

L 

16 

X 

3.  125 

50.0 

1 

51,962.5 

1 1 
H = H = 

0.05233 

lbs. 

R L 


•.fin  • 


■ m i ^-ifcatoea 


I 

f 


r 


I 


>j 


•r 


•ar 


J 


I 

/ 


, / 


T. 


' i V 


\ 


-8S- 


Case 


Case 


B: 


d = 460,907 
B 


V = V = 25  lbs. 
R L 


143,269.8 
0.3108  lbs. 

0.6892  lbs. 


d 

V 


1 

1 ^ 
R 

v?  = 


71,075.2 
0.  1542  lbs. 

0.8458  lbs, 


C: 


d + ^ = 193.77 

K 4 


J- 

0.079 

X 

3. 125 

0.25 

2 

4.55 

X 

6.25 

=s 

28.44 

3 

5.80 

X 

6.25 

36.25 

6 

6.  33 

X 

6.25 

39.56 

7 

5.21 

X 

6.25 

=: 

32.56 

10 

5.40 

X 

6.25 

s 

33.75 

11 

4.70 

X 

6.25 

s 

29.38 

13 

3.40 

X 

6.25 

=: 

21.25 

L 

0.048 

X 

3. 125 

0.15 

221.59 

^ ^ ^ ^ 8 SI. 59  X 180  ^ 205.84  in.  lbs. 
R L 193.77 


d =6.12 

4 


"r 


= 5.685  in.  lbs. 


M = 4.665  in.  lbs. 
L 


di=  5-20 

= -4.8  30  in.  lbs. 

R 

= 3.600  in.  lbs. 
L 


Y = Y = ,805.,.84  = 81.76  in. 

^ ^ 2.5177 


= 90.25  in. 
R 

= 74.06  in. 
L 


Y = 92. 30  in. 
R 

Yl  = 68.79  in. 


I 


u 


'■± 


/: 


1 


.■s 


i 


—84— 


Case  A: 


Span  50'  - Col,  Height  26* 
t = 38.5  ft. 

Class  I. 


= 164,564 


J 

33 

X 

3. 125 

=: 

103, 13 

H = H = 

2 

18  35 

X 

6.25 

= 

1], 468. 75 

R L 

3 

2375 

X 

6.25 

= 

14,843.75 

6 

2660 

X 

6,25 

=: 

16,625.00 

d - 

7 

2240 

X 

6.  25 

= 

14,000.00 

4 

10 

2660 

X 

6.25 

16, 625 . 00 

"h  = fL : 

11 

2375 

X 

6,25 

— 

14,843.75 

13 

18  35 

X 

6.25 

- 

11,468.75 

L 

33 

X 

3. 125 

103. 13 
100,081.26 

”r  t 

= 0.6082  lbs. 


2*475 


0.01504  lbs. 
= 2,080 

^ 0.01264  lbs. 


Case  B: 


V*  = 


V = V 
R L 

0.3125  lbs, 
0.6875  lbs. 


= 25  lbs. 


V = 
R 

1 


0.15625  lbs. 
0.84375  lbs. 


Class  II 


Case  A: 


d ~ 42,234 


J: 

19 

X 

3.  125 

= 59.38 

2: 

1100 

X 

6.25 

= 6,875.00 

3: 

1460 

X 

6.25 

= 9,125.00 

6: 

1600 

X 

6.25 

=10,000.00 

7: 

1400 

X 

6.25 

= 8,750.00 

10: 

1600 

X 

6.25 

=10,000.00 

11: 

1460 

X 

6.25 

= 9,125.00 

13: 

1100 

X 

6.25 

= 6,875.00 

L: 

19 

X 

3. 125 

= 59.38 

60,868.76 

R 


L 


1.4412  lbs. 
1,556 

0.03684  lbs. 
1,276 

0.03021  lbs. 


V 


( 


I 


•' 


• ^ • 
r 


S 


r 

’u- 

1 


. 


.11^  »r;.  '..,i 


^ O 'k  ■ 


Case 


Case 


B:  d =*  593,912 

B 

Y =s  V — 25  lb  s . 
R L 

(i  = 184,834S 


= 0.3112  lbs. 
R 

= 0.6888  lbs, 
L 


d-,  = 91,857.8 
1 

= 0.  1547  lbs. 
R 

= 0.8453  lbs. 
L 


C: 

d + 

4 = 325.81 

K 

4 

J: 

0.075‘x 

3.125 

= 0.23 

2: 

4.28  X 

6.25 

= 26.75 

3: 

5.45  X 

6.25 

= 34.06 

6; 

5.83  X 

6.25 

= 36.44 

7: 

4.75  X 

6.25 

= 29.69 

5 

10: 

5 . 00  X 

6.25 

= 31.25 

11: 

4.40  X 

6.25 

= 27.50 

13: 

3.10  X 

6.25 

= 19.37 

L: 

0.045X 

3.125 

= 0.14 

205 . 43 

M =: 

M = 

205.43  X 240 

=:  151.33  in. 

lbs. 

R 

L 

325.81 

d 

= 5.68 

d =4".  90 
1 

4 

“r 

= 4.184 

in.  lbs. 

=s3.  609 
R 

in. 

“l 

= 3.404 

in.  lbs. 

=2.679 

Li 

in. 

=:  Y 

T 

= 151.33 

105.00  in. 

Li 

1,4412 

= 113.57  in. 
R 

y4  = 92.40  in. 

L 


1 

= 119.46  in. 

1 

Y = 88.68  in. 

L 


-86- 


Span  50'  - Col.  Height  ol' 
t = 45.5  ft. 

Class  I. 


Case  A: 

= 124,285 

J: 

bb  X 

3.125 

= 10b. 13 

= H 

= 0.9483  lbs. 

2: 

2175  X 

6.25 

= 13,593.75 

R 

L 

5: 

2800  X 

6.25 

= 17,500.00 

d 

= 3,000 

ol 

blOO  X 

6. 35 

= 19,375.00 

4 

7: 

2675  X 

6.25 

= 16,718.75 

4 

4 

10: 

blOO  X 

6,  25 

= 19,375.00 

H 

= H 

= 0.02414  lbs 

11: 

2800  X 

6,25 

= 17,500.00 

R 

L 

15: 

2175  X 

6.25 

= 13,593.75 

d. 

= 2,450 

L: 

bb  X 

3.125 

= 103.13 

117,862.51 

1 

= «£ 

= 0,01972  lbs 

Case  3: 

V = V =25  lbs. 

R 

L 

= 0.bl25  lbs. 

v' 

= 0. 

15625  lbs. 

R 

R 

= 0.6675  lbs. 

J. 

V 

= 0. 

84375  lbs. 

L 

Class  II 

L 

Case  A: 

d 

B 

= 32,569 

J: 

17  X 

b.125 

= 53.13 

H = 

H 

= 2.1094  lbs. 

0 • 

1236  X 

6.25 

= 7 ,725.00 

R 

L 

b : 

1640  X 

6.25 

=10,250.00 

d 

= 1,755 

r*  - 

D : 

1605  X 

6 . 25 

=11,281.25 

4 

7: 

10: 

1620  X 
1805  X 

6.25 

6.25 

=10,125.00 

=11,281.25 

h‘= 

= 0.0538  lbs. 

11: 

1640  X 

6.25 

=10,250.00 

R 

L 

loi 

L: 

1236  x~ 
17  X 

6,25 

3.125 

= 7,725,00 
= 53.13 

I 

H = 

d 

= 1,410 

68,743.76 

= 0.04327  lbs. 

R 

L 

J 


f.,  • 


t 


V- 


/ 


-87- 


Case  B: 


a 


B 


271,377 


= V = 25  lbs. 

R L 

d = 84,045.9 


= 0.3097  lbs. 

R 

=:  0.6903  lbs. 

L 

Case  C:  d + ^ = 190.97 


J 

0.053 

X 

2 

3.60 

X 

3 

4.65 

X 

6 

4.93 

X 

7 

3.90 

X 

10 

4.01 

X 

11 

3.40 

X 

13 

2.52 

X 

L 

0.02S 

X 

3.125  = 0.17 

6.25  = 22.50 

6.25  = 29.06 

6.25  = 30.81 

6.25  = 24.38 

6.25  = 25.06 

6.25  = 21.25 

6.25  = 15.75 

3.125  ^ ,09 

169.07 


= 41,463.3 

= 0. 1528  lbs. 
R 
1 

V,  = 0.8472  lbs. 

Li 


M = M = 169.07  X 300  = 265.60  in.  lbs. 

R L 190.97 


d ==  4.80  = 4.20 

4 


4 

7.540 

in. 

lbs. 

“r'  = 

6. 

598 

in. 

lbs. 

“r 

5.860 

in. 

lbs. 

L 

4. 

528 

in. 

lbs. 

Y 

= 265.60 

= 125.91  in. 

R 

L 

2. 1094 

II 

>-• 

140.02 

it^. 

II 

15 

2.48 

in. 

R 

R 

y4  = 

108.82 

in. 

1 

104.65 

in. 

L 


, ‘ - - • . *t--- 

I ’ 


-88- 


Span  60  ' - Col.  Height  16  ' 
Total  Vertical  Load  on  Truss  = 60  lbs. 
t = 31  ft. 


Class  I. 


Case  A: 


= 28,327 


J 

17 

X 

3.75 

63.75 

H = 

T 

2.5383  lbs. 

2 

1090 

X 

7.5 

s 

8,175.00 

K 

L 

3 

1420 

X 

7.5 

3 

10,650.00 

A ^ 

1, 485 

6 

1580 

X 

7.5 

S 

11,850.00 

a 

4 

7 

1390 

X 

7.5 

= 

10,425.00 

10 

1580 

X 

7.5 

S 

11,850.00 

= 

0.05242  lbs 

11 

1420 

X 

7.5 

= 

10,650.00 

R 

L 

13 

1090 

X 

7.5 

8, 175.00 

L 

17 

X 

3.75 

63.75 

1,244 

71,902.50 

1 

H = 

= 

0.04392  lbs 

R 


Case  B: 


-R  = 
V*  = 


V = V = 30  lbs. 
R L 

1 X 225  =r  0.3125  lbs. 

720 

1 X 495  = 0,6875  lbs. 

720 

Class  II. 


V = 1 X 113.5  0.15625ltee. 

0.843751bs. 


V 


R 

1 


= 1 


720 

X 607.5 


720 


e A? 

d 

3 

7,868 

J 

10 

X 

B 

3.75 

37.5 

2 

745 

X 

7.5 

S 

5,587.5 

3 

990 

X 

7.5 

= 

7,425.0 

6 

1105 

X 

7.5 

S 

8,287.5 

7 

1010 

X 

7.5 

7,575.0 

10 

1105 

X 

7.5 

=; 

8,287.5 

11 

990 

X 

7.5 

— 

7,245.0 

13 

745 

X 

7.5 

= 

5.587.5 

L 

10 

X 

3.75 

37.5 

50,070.0 


fj  =s  H — 6. 3638  lbs. 
R L 

d = 1, 060 


13472  lbs. 
= 844 


H = H = 

R L 


ai0727  lbs 


Ak 


t 


I 


T 


^ •• 


/ 


J 

V 


( 


'.-i 


-89' 


Case 


C ase 


B:  d = 439,954 

B 


Y = V - ^0  lbs. 
R L 

d = 146,062.5 

y*  = 0.5108  lbs. 

R 

= 0.6892  lbs. 

L 


C: 

d + 
K 

4 

= 

95. 

J: 

0.072 

X 

5.75 

0.27 

2: 

4.90 

X 

7.5 

= 

56.75 

3: 

6.55 

X 

7.5 

49.13 

6: 

7.08 

X 

7.5 

= 

55.10 

7: 

6.10 

X 

7.5 

45,75 

10: 

3.41 

X 

7.5 

= 

48.07 

11: 

5.45 

X 

7.5 

=5 

40.88 

15: 

5. 85 

X 

7.5 

= 

28.87 

L: 

0.046 

x3L7  5 

=: 

0.17 

302.99 


Mn  = M =-b02.99  X 130  = 380.72 

R L 95 . 50 


d =6,89 


= 8.657  in.  lbs. 

R 

= 7.455  in.  lbs. 

L 

Y = Y = 580 . 72 
^ ^ 3.5658 

= 64.26  in. 

R 

y"*”  , = 55.17  in. 

L 


1 

1 

Li 

L 

58.85  in. 


= 72,425.1 
= 0.1541  lbs. 
= 0.8459  lbs. 


in.  lbs. 

= 5.69 

= 7.150  in,  lbs. 

= 5.602  in.  lbs. 

= 66,65  in. 

= 52.22  in. 


-90- 


Case 


Case 


Case 


Span 

60*  - Col. 

Height 

21* 

t s 36  ft. 

Class  I. 

A: 

V 

76, 182 

J 

22 

X 

3.75  = 

82.5 

H = 

H = 

1.1919 

lbs. 

2 

1360 

X 

7.5  = 

10,200.0 

R 

L 

3 

1815 

X 

7.5  = 

13,612.5 

1,885 

6 

2000 

X 

7.5  = 

15,000.0 

d = 

4 

7 

1735 

X 

7.5  = 

13,012.5 

4 

% = 

= 

0.02474 

lbs 

10 

2000 

X 

7.5  = 

15,000.0 

11 

1315 

X 

7.5  = 

13,612.5 

13 

1360 

X 

7.5  =: 

10,200.0 

di“ 

1,570 

L 

22 

X 

3.75  = 

82.5 

1 

i 

0.02061 

lbs 

90,802.5 

II 

B: 

V 

= V 

= 30  lbs. 

R L 

1 

0 

o 

II 

> 

5125  lbs. 

0.15625  lbs. 

‘ = 0.6875  lbs. 

L 

1 

0.84375  lbs. 

Class  II, 

A: 

d 

19,852 

B 

J 

13 

X 

3.75  = 

48.75 

H = 

H = 

2,8610 

lbs. 

2 

850 

X 

7.5 

6,375.00 

R 

L 

3 

1130 

X 

7.5  = 

8,475.00 

6 

1250 

X 

7.5 

9, 375.00 

d “ 

1,190 

7 

1100 

X 

7.5  = 

8,250.00 

4 

10 

1250 

X 

7.5  = 

9, 375.00 

0.05994 

lbs 

11 

1130 

X 

7.5  = 

8,475.00 

H4  = 

13 

850 

X 

7.5  = 

6,375.00 

R 

L 

980 

L 

13 

X 

3.75  = 

48.75 

56,797.50 

1 = 

= 

0.04937 

lbs 

R L 


I 


I 


't 

1 


i 'J'i 


Case  B:  d ~ 681,481 

B 

V = V = bO  lbs. 
R L 

d = 205,912.5 


= O.ollb  lbs. 
R 

= 0.6887  lbs. 
L 


Case  C: 

d 

K 

+ 

4 

rz 

191.18 

J: 

0.038 

X 

3.75 

= 

0.26 

2: 

4.33 

X 

7.5 

32.40 

3: 

5.76 

X 

7.5 

= 

4o . 35 

6: 

S.15 

X 

7.5 

46.12 

7: 

5.39 

X 

7.5 

40.43 

10: 

5.70 

X 

7.5 

= 

42.75 

11: 

5.10 

X 

7.5 

=: 

38 . 35 

13: 

3.78 

X 

7.5 

26.35 

L: 

0.043 

X 

3.75 

= 0.16 
272.07 

-91- 


102,b50.1 
0.1547  lbs. 
0.845O  lbs. 


Y = M = 272.07  X 180  = 253.16  in.  lbs. 

R L ' 151.18- 


d = S.OO  d = 5.11 

4 ^ 


“r 

= 5.649  in.  lbs. 

= 4.811 

in. 

lbs. 

L 

= 4.785  in.  lbs. 

= 3.695 

in. 

lbs. 

Y = Y =356.16 

R L 2.861 

= 89.54  in. 

"r 

= 94. 24  in. 

R 

= 97.45 

in. 

T 

= 79.83  in. 

1 

Y 

= 74.84 

in. 

L 


A 


( 

] 


I 

i: 


) 


V 


» 


Span  60'.  - Col.  Keigiit  26’ 
t=  41  ft. 

Class  I. 


Case  A: 


= 131,137 


B 


J: 

28 

X 

0.75 

= 

105 

II 

= 0.6809  lbs. 

2: 

1640 

X 

7.5 

= 

12,500 

R L 

5: 

2160 

X 

7.5 

= 

13,200 

d 

= 2,520 

6: 

2450 

X 

7.5 

18,225 

4 

7: 

2140 

X 

7.5 

=: 

16,050 

10: 

2450 

X 

7.5 

18,225 

K 

= 0.01440  lbs. 

11: 

2160 

X 

7.5 

= 

13 , 200 

R L 

15: 

1640 

X 

7.5 

12,500 

= 1,920 

L: 

28 

X 

5.75 

= 

105 

1 

109,710 


Hd  = H = 0.01192  lbs, 

R T. 


Case  B: 


V = V = 50  lbs. 

4 R L 

V =0.5125  lbs. 

R 


1 

V = 0.15625  lbs. 
R 


V""  = 0.3875  lbs, 

L 


= 0.64575  lbs. 
L 


Class  II. 


Case  A: 

d 

B 

= 

41,105 

J: 

16 

X 

5.75 

60 

H = H = 1.64o5  lbs. 

2: 

1000 

X 

7.5 

= 

7,500 

R L 

‘A  • 

w • 

6: 

1540 

1500 

X 

X 

7.5 

7.5 

= 

10,050 

11,250 

d = 1,456 

7: 

1510 

X 

7.5 

= 

9,825 

4 

10: 

11: 

1500 

1540 

X 

X 

7.5 

7.5 

= 

11,250 

1C,G6'0 

= 0.05499  lbs 

R L 

-15: 

1000 

X 

7.5 

= 

7,500 

L: 

16 

X 

5.75 

= 

60 

a = 1,176 

67,545 

1 ' 

= 0.02861  lbs. 

R L 


* 


A 

■( 

* ' _ 

iiola- 

V;  “'J*  * . 


-94- 


Span  60'  - Col.  Height  61' 


t = 46  ft. 
Cl  as  s I • 


Case  A:  d = 119,448 


B 


J: 

27 

X 

3,75 

= 

101.25 

^ • 

1925 

X 

7.5 

14,437.50 

3 : 

2550 

X 

7.5 

= 

19,125.00 

6: 

2850 

X 

7.5 

= 

21,375.00 

7: 

2525 

X 

7.5 

=r 

16,937.50 

10: 

2850 

X 

7.5 

= 

21,375.00 

11: 

2550 

X 

7.5 

=: 

18,125.00 

13: 

1925 

X 

7.5 

= 

14,437.50 

L: 

27 

X 

3.75 

= 

101.25 

129,015. 


H = H = 1.0801  lbs. 
R L 

d = 2,750 


= 0.02302  lbs 

R L 

d = 2,250 
1 

= 0.01884  lbs 

R L 


Case  B: 


4 


V 


4 


L 


V 

R 


0.3125  lbs. 
0.6875  lbs. 


V = 30  lbs. 

Li 


y = 0. 15625  lbs. 

'r 

= 0.84  375  lbs. 
L 


Class  II. 


Case  ffl:  ^ 

B 


J: 

15 

X 

2 : 

1120 

X 

3: 

1480 

X 

6: 

1630 

X 

7: 

1420 

X 

10: 

1630 

X 

11: 

1480 

X 

13: 

1120 

X 

L: 

15 

X 

= 31,150 

3.75  = 56.25 

7.5  = 8,400.00 

7.5  =11,100.00 

7.5  =12,225.00 

7.5  =10,650.00 

7.5  =12,225.00 

7.5  =11,100.00 

7.5  = 8,400.00 

3.75  = 56.35 

74,212.50 


2.3824  lbs. 
1,570 

0. 05040  lbs 

1 , c oO 

0.04270  lbs 


I 


-95' 


Case 


Case 


B: 

d 

: 388,565 

B 

II 

> 

II 

( 

> 

d 

= 120,631 

.3  « I' 

4 

4 

V 

= 0.3105 

lbs . 

R 

4 

V 

= 0.6895 

lbs. 

L 

C: 

d + 

II 

I-* 

OJ 

CD 

K 

4 

J: 

0.046 

X 

3.75  = 0.17 

O • 
• 

3.35 

X 

7.5  =25.13 

3: 

4.34 

X 

7.5  = 32.55 

6: 

4.67 

X 

7.5  = 35.02 

7: 

3. 85 

X 

7.5  = 28.88 

10: 

4.09 

X 

7.5  = 30.67 

11: 

3.55 

X 

7.5  = 23.63 

13: 

2.65 

X 

7.5  = 19.87 

L: 

0.025 

X 

3.75  = .09 

199.01 


M = M = 199.01  X ;500 
^ ^ 188.27 


lbs , 

d 

1 


R 

1 

V 

L 


= 317.11 


d = 4.53 


= 7.218  in.  lbs. 

= 5.778  in.  lbs. 
B 


M 


m: 


M = Y = 317.11 
R L 2.3824 

Y4  =:  143.21  in. 

R 

Y^  = 114.64  in. 

L> 


133.11  in. 


1 

Y 

R 

1 

Y 

L 


59,709.5 
0.1537  lbs. 
0.8463  lbs. 


in.  lbs. 

^ 3.94 

6. 276  in.  lbs. 
^ 4.442  in.  lbs. 

: 147.03  in. 
104.03  in. 


/ 


I 


96 


TABLE  1. 


DESIGN  OE  BENT 
SPAN  LENGTH  40  PT, 


A 

L 

1 

Member 

* Section 

Area 

in  sq.in. 

Length  in 

in.  A 

G 

J 

Same  as  column 

1/4, 

7.24 

72.00 

9.9448 

J 

1 

2 Ls  2 1/4x2  1/4  x 

2.12 

75.00 

35.3773 

G 

1 / 

•?2  is  2 3/4  X 2 3/4 

X 1/4 

2.12 

103.97 

49.0424 

J 

2 1/4x2  1/4  X 

2^ 

1/4 

2.62 

67.08 

25.6031 

1 

1.06 

33.64 

31.6415 

2 

3 

2.62 

67.08 

26.6031 

1 

3 

2.12 

75.00 

35.3773 

1 

4 

2.12 

75.00 

35.3773 

3 

4 

2.12 

67.08 

31.6415 

3 

5 

1.06 

75.00 

70.7547 

4 

5 

1.06 

75.00 

70.7547 

3 

6 

All  members  having 

the 

2.62 

6 7.08 

25.6031 

6 

6 

1.06 

33.54 

31.6415 

6 

7 

same  area  have  the 

same 

2.62 

67.08 

25.6031 

5 

7 

1.06 

75.00 

70.7547 

4 

8 

section 

2.12 

180.00 

84.9057 

7 

9 

1.06 

75.00 

70.7547 

7 

10 

2.62 

67.08 

26.6031 

9 

10 

1.06 

33.54 

31.6415 

10 

11 

2.62 

67.08 

25.6031 

8 

9 

1.06 

75.00 

70.7547 

9 

11 

1.06 

76.00 

70.7547 

8 

11 

2.12 

67.08 

31.6415 

8 

12 

2.12 

75.00 

35.3773 

11 

12 

2.12 

76.00 

36.3773 

11 

13 

2.62 

67.08 

26.6031 

12 

13 

1.06 

33.64 

31.6415 

13 

L 

2.62 

67.08 

25.6031 

12 

Z 

2.12 

103.97 

49.0424 

12 

I 

2.12 

75.00 

35.3773 

K 

I 

7.24 

72.00 

9.9448 

For  the  16,21,  and  26-ft,  coltiran  heights  the  columns 
D J and  B L are  made  of  1 PI.  8 x l/4  and  ^ L3  3x2l/2x  1/4 
with  the  longer  leg  outstanding.  The  moment  of  inertia  of  the 
column  section  about  a vertical  axis  normal  to  the  plane  of  the 
truss  is  81.2  inches.^  The  moment  of  inertia  for  the  31- foot 
column  height  is  222.0  in.^. 

♦ 


See  Pig.  1 


M . 


I 


X . . 
.1 


- * J- 


...  J. 

‘.'.I. 


X.. 


TABLE  2 - 

COI^UTATION  OP  ^ 

A 

SPAU  40»  - Col.Hgt.  21* 


Class  I,  Case  A 

Class  II, 

Case  A 

Member 

* 

hL 

Ul 

Pi  Ull 

Pg  = s 

X P + Q 

..  . A. 

A 

s 

J 

-3.370 

- 33.514 

-0.8100 

+ 27.415 

-0.9600 

-0.9600 

J 

1 

+4.230 

+149.64  6 

+0.6200 

+ 93.529 

+0.9200 

+1.9200 

Or 

1 

+4.860 

+238.346 

+1.3862 

+332.493 

+1.3862 

+1.3862 

J 

2 

-7.496 

-191.89  6 

-1.8112 

+349.248 

-2.1466 

-2.1466 

2 

3 

-7.495 

-191.895 

-1.8112 

+349.248 

-2.1466 

-2.1466 

1 

3 

+4.220 

+149.292 

+1.2000 

+180.793 

+1.2000 

+1.2000 

1 

4 

+5.196 

+183.786 

+0.9000 

+164.488 

+1.2000 

+2.2000 

3 

4 

-1.875 

-69.328 

-0.5367 

+31.859 

-0.5367 

-0.5367 

4 

6 

+2.100 

+148.586 

+0.6000 

+89.151 

+0.6000 

+0.6000 

3 

6 

-3.745 

-95.884 

-0.7379 

+70.187 

-1.0733 

-1.0733 

6 

7 

-3.745 

-95.884 

-0.7379 

+70.  187 

-1.0733 

-1.0733 

5 

7 

+2.100 

+148.686 

+0.6000 

+89.151 

+0.6000 

+0.6000 

4 

8 

+3.100 

+263.208 

+0.3000 

+76.857 

+0.6000 

+1.6000 

7 

9 

+2.100 

+148.686 

+0.6000 

+0.6000 

7 

10 

-3.745 

-96.884 

-0.3364 

+31.642 

-1.0733 

-1.0733 

10 

11 

-5.475 

—96.884 

-0.3364 

+31.642 

-1.0733 

-1.0733 

8 

9 

+2.100 

+148.585 

+0.6000 

+0.6000 

8 

11 

-1.876 

- 59.328 

-0.5367 

-0.5367 

8 

12 

+6.195 

+183.785 

+0.3000 

+53.665 

+1.2000 

+2.2000 

11 

12 

+4. 220 

+149.292 

+1.2000 

+1.2000 

11 

13 

-7.49  5 

-191.896 

-0.3354 

+ 63.325 

-2.1466 

-2.1466 

13 

I 

-7.496 

-191.895 

-0.3354 

+63.325 

-2.1466 

-2.1466 

12 

E 

+4.860 

+238.346 

+1.3862 

+1.3862 

12 

L 

+4.230 

+149.646 

+0.3000 

+43.697 

+0.9200 

+1.9200 

K 

I 

-3.370 

-33.514 

-0.1500 

+5.027 

-0.9600 

-0.9600 

+2216.9^7 


A,JJj 


930.070 

E 


inches 


_ 2216.927 

^ A E ^ E 


inches 


*See  Pig.  1. 


98, 


TABLE  2 (continued) 


Class  II,  Case 

A 

Member* 

A 

A A 

I’e  Pi  t _ 

A 

SPU,i  , 

A A 

Pg  1 

A 

iiL. 

A 

A 

A 

A 

G 

J 

- 9.5470 

- 9.5470 

+ 7.7331 

+ 7.7331 

- 19.05 

J 

1 

+32.5471 

+ 67.9244 

+20.1792 

+42.1131 

+100.19 

G 

1 

+67.9826 

+ 67.9826 

+94.2375 

+94.2375 

+136.36 

J 

2 

-64.9596 

- 54.9596 

+99.5428 

+99.5428 

-109.58 

2 

3 

-54.9696 

- 54.9596 

+99.5428 

+99.5428 

-109.58 

1 

3 

+42.4528 

+ 42.4528 

+50.9434 

+50.9434 

+84.91 

1 

4 

+42.4528 

+ 77.8301 

+38.2075 

+70.0471 

+120.28 

3 

4 

-16.9820 

- 16.9820 

+ 9.1142 

+ 9.1142 

-33.54 

4 

5 

+42.4528 

+ 42.4528 

+25.4717 

+ 25.4717 

+84.  91 

3 

6 

-27.4798 

- 27.4798 

+20.2773 

+20.2773 

-54.79 

6 

7 

-27.4798 

- 27.4798 

+20.2773 

+20.2773 

-54.79 

6 

7 

+42.4528 

+ 42.4528 

+26.4717 

+26.4717 

+84.91 

4 

8 

+50.9434 

+135.8491 

+15.2830 

+40.7547 

+186.79 

7 

9 

+42.4528 

+ 42.4528 

+84.91 

7 

10 

-27.4798 

- 27.4798 

+ 9.2167 

+ 9.2167 

-54.79 

10 

11 

-27.4798 

- 27.4798 

+ 9.2167 

+ 9.2167 

-54.79 

8 

9 

+42.4528 

+42.  4528 

+84.91 

8 

11 

-16.9820 

- 16.9820 

-33.54 

.8 

12 

+42.4628 

+ 77.8301 

+12.7358 

+23.3490 

+120.28 

11 

12 

+42.4528 

+ 42.4528 

+84.91 

11 

13 

-64.969  6 

-54  .9596 

+18.4334 

+18.4334 

-109.68 

13 

L 

-54.9596 

- 54.9596 

+18.4334 

+18.4334 

-109.68 

12 

K 

467.9826 

+67.  9826 

+135.36 

12 

L 

+32.5471 

+ 67  . 9 244 

+ 9.7641 

+20.3773 

+100.19 

K 

L 

- 9.5470 

- 9.5470 

+ 1.4321 

+ 1.4321 

-19.05 

+605.5137 

+70^.9853 

, „ 200.9432P  + 427.3681  . , 

= g inches 

605.5137?  7p5..985a 

^ — AE  B 


*See  Pig.  1 


TABI.E  3 

COMPUTATION  OP  P ^ 

A 

SPAN  40'  - COL,  HGT,  21* 


99 


pq 

(D 

CO 

03 

o 


OQ 

(Q 

d 

H 

O 


+ 

to 

o 

'-y 

+ 

to 

t=> 

CO 

n 

to 

t=3 

to 

(U 


to 

t=> 


»s) 

E^ 


i 


CO 


L 


^1- 


EH 

+ 

O 

M 

cf 

CO 

'k 


* 

V 

1 


to 

t3 

EH 


<vJ 

CO 

tJ 


Csj 

to 

CO 


Clj 


sh 

raiil 


EH 


<3? 


CO 


UD  to 

*>  l> 


to  ^ 
+ + 
CVJ  C-  H H 
H Nt* 

H t-  C7>  0> 
H ^ ^ 

• • • • 

O)  LO  o> 

W CV2  W 

+ + + + 

CM  C-  H H 
H 

f— 1 C7»  cy> 
iH  ^ vj<  tj< 

• • • • 

0>  lO  CJ)  O) 
CM  CM  CM 
+ + + + 


CM  to  CM  CM 
00  iH  "4' 

CJ>  H 00  00 
M5  CD  H H 
(3t 

+ + 

l>-  tO  i>- 
rH  CD  rH 
t>  lO  I> 

to 

• • • 

lO  !>-  LO 
CM  H CM 
+ CM  + 

*1“ 

t>-  t~  iH  rH 
H O'  H ^ 
t-  O IS  O' 
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• • • • 

to  H W 

CM  CD  CM 

+ + + 


iQ  H 
H O CM 
H lO  CD  CM 
CD  IS  CJ>  ^ 


to  to  to  O 

to  vO  t-  CM 
CM  CM  !p 

to  to  CM 

> 


CsOtOWOOOlOCOoC'OOcOtOCMO 

tDOCOWOOOCOtOO^OOM^tOcOo 

tOOIStsOoOSsISOt^OO'^'sliOOCM 

lOM500cDtOMOOOcOinCMCMrH|HtOC3' 


o 

H 

H CM 

CM 

H CM 

O O H 

H 

1 

+ 

+ 1 

1 

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1 + 1 

1 

00 

<D 

H 

to 

H 

H 

H 

H 

H 

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to 

tQ 

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IQ 

IQ 

IQ 

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CM 

IQ 

IS 

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• 

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1 

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1 

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i 

1 

+ + + I I + I + + 

^QOO'O'OOOOOO^lt 

HQOrHHOOtOlOcD 

HHlOiOHO^H 

OOOCJ'O'OCMCMO 

• •«••••« 

CTitOlOlOlOtOCMlO 

ooocmcmcdHhoo 

iHCMHHCMtHCMCM 

+ + + + + • + + 


CM  CM  H H 

•I+  + 

CO  to  lO  o 
CM  CM  IQ  to  I 
•Q  m iH  CM  ' 
rH  iH  CM  CM 

• • • • 

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O O iQ  'si* 

to  w ^ H 


a 


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as 

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H H 

o 

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to  to 

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to  to 

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to  IQ 

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CM  CM 

o 

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• • 

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CM  CM 

CM 

CM  CM 

® + 

• 1 

+ 

1 1 

OO-tJt-stiOISOoOOOOtOtOO 

000'0'OJsOO'<^'^HtOO 

OOHHOlsOOl>IS'ti»tOO 

OO0'CJ>Ol0OOOOCM|H^ 

• •••••••••••• 

CMtl<'d*'ti<'>tttlOMDOOCMCMO'-«;Jt‘Q 
+ +I  | + I+  +rHi^++l 
I I 


CM 

§m  o 

05 

CO  Cf 
CM 

pq 


o 

o5  o5 


OQ  O 


CO  CO 


OH  H CM  CM  to 

i-aHHCMlOCO’tlt’.^iOtOC'-lscDO'HHO'HHHHf^MHlH 

O CD  rH  H to  CM  CM  M 
O O ►;>  CM  H H to CO  MO  to '«;J<  t- IS  H CD  00  H H H H H 


T 

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1 


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1 

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1 


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TABLE  3 (oontinued) 


100. 


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0) 

CQ 

o5 

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M 

M 

(0 

CQ 

CtJ 

rH 

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P4 


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i, 

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t 


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V.  f 

>*./• 

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102 


Eh 

Eh  O 
S CV2 


lopq  g 

Eh  M 
W ^ 
M Ph 
P CO 


M<il 


GO 

lO 

c- 

o 

H* 

o 

CVJ 

o 

CVJ 

o 

E> 

lO 

00 

CVJ 

to 

rH 

t<3 

to 

to 

CO 

CVJ 

CO 

to 

rH 

CO 

CVJ 

LO 

lO 

o 

'd* 

o 

>d< 

1 1 

H* 

iH 

00 

o 

to 

to 

'd* 

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cvj 

to 

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fH 

o 

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to 

CVJ 

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00 

to 

00 

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to 

to 

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to 

to 

to 

to 

to 

to 

00 

to 

CVJ 

o 

to 

c» 

CO 

rH 

CVl 

H 

to 

o 

CVJ 

CO 

rH 

to 

to 

CO 

'd* 

GO 

«>- 

rH 

c- 

to 

to 

CO 

iH 

rH 

CVJ 

CVJ 

CO 

CVJ 

CO 

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c- 

CVJ 

CVJ 

rH 

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rH 

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t> 

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lO 

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to 

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to 

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to 

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to 

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00 

00 

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tjC 

Si 

(D 

P 


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0 

0 


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O^tO  COOOWtocvj^OCOtOCMCDCVlvO 
00<MHtO«30tOOtOo^COHC\l 

• •••••••••••• 

C0WCV2<MHcvJH<MHO2WWC0 


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H 

W rH  I — 1 - 

^ M M H 
M CV2  W M 

CV2  H M 

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M rH  rH  Cvj*^ 

O M H 

rH  to  (M  to 

CQ 


I — 1 

M 

■>;}«  CVl 

M iH  iH 

w MW  M 

■ — .CV3  M w 

H M w M 


^ H H 

to  <M 

W cvj 

CQ  CQ 

PPP  P 


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H H H 

M M 

CVl  CVJ  CVJ 

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rH  CvJ  CVJ 
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to  CVJ  CVJ 
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© 

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d 

p 

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a 

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Si 

O 

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§ 

03 

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rH 

ra  »d 

d 

CVJ 

S ^ 

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P 

S cd 

p 

00 

p p 

P 

H 03 

© 

03 

o 

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p 

bo 

P © 

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d 

A CVJ 

P 

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A to 

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03 

p 

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P rH 

JH 

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a 

p 

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p 

1 

p 

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• CVJ 

<H  rH 

I 

to  CVJ 

H M 
CVJ 
'H" — . 
Si  H 
d 

to 

• 

CQ 

«H  iA 

CJ  ^ 
H 
-txd 
■P  Si 
'H  cd 


I 

O 

rH 

0> 

A 

P 


'id* 

iH 

M 

o 

I — I 


^ H 
p P 

Ph 

H 

«H 

O 


O 

o 

<D  CQ 

rC|  © 

o 

*p  si 

O P 

oj  »0 

P to 

p p 
>H 

© 03 
Si  *H 


tp 

O 

P 

SI 

© 

s 

o 

S 

© 

g 


ca 

03 

pi 

P 

p 

© 

p 

<P 

o 

© 

SJ 

cd 

H 

Pr 


© 

P 

P 

O 

P 

d 

•H 

P 

P 

© 

S! 

•H 

«H 

O 

P 

Si 

© 

s 

o 

S 

© 


See  Fig,  22 


103 


TABLE  6 

DESIGN  OE  BENT 
SPAN  LENGTH  30  ft. 


A 

Member*  Section  Area  Sq..  In. 

Length  in, 

L 

A 

• -ri 

S J 

Same  as  columns  7,24 

48.000 

6.6298 

J 1 

2l3  2 l/2x2xl/4  2.12 

56.250 

26.5330 

G 1 

2.12 

73.946 

34,8802 

J 2 

2/52  l/2x  21/2x1/4  2,38 

50.312 

21,1395 

1 2 

IL  2 1/2x2  xl/4  1.06 

25,156 

23.7321 

2 3 

2.38 

50.312 

21.1395 

1 3 

2.12 

56.250 

26.5330 

1 4 

2.12 

56.250 

26.5330 

3 4 

All  members  having  the  2,12 

50.312 

23.7321 

3 5 

1.06 

56,250 

53.0660 

4 5 

same  area  have  the  same  1,06 

56,250 

53,0660 

3 6 

2.38 

50.312 

21.1395 

5 6 

section  1.06 

25.156 

23.7321 

6 7 

2.38 

50.312 

21.1395 

5 7 

1.06 

56.250 

53.0660 

4 8 
7 9 

7 10 
9 10 
1011 

8 9 

9 11 

2.12 

135.000 

63.6792 

8 11 
8 12 
1112 
1113 
1213 
13  L 
12  K 
12  L 
E 1 

Symmetrical  with  the 

other  half 

of  Truss 

For  the  16-f t. , 21-ft. , and  E6-ft.  column  heights  the  coltunns  DJ 
and  BL  are  made  of  1 PI.  8 x l/4  and  4t  L3  3x2l/2x  l/4  with  the 
longer  leg  outstanding.  The  moment  of  inertia  of  the  column 


section  about  a vertical  axis  normal  to  the  plane  of  the  truss  is 
81,2  inches^. 

The  moment  of  inertia  for  the  31- foot  column  height  is  222,0  in^ 

*See  Pig.  23. 


r’l/iTiilW 


«■ 

/f 


\ 


> r 


j 


I 


f r 


. ''I  Q 


.rJ5j:isar 


TABLE  7 


104. 


DESIGN  OF  BENT 
SPAN  LENGTH  50  ft. 


Member^  Section  A 

Area  Sq.  in 

1 

. ieneth  in 

z. 

in.  . A 

G 

J Same  as  columns. 

7.24 

72.000 

9.9448 

J 

1 2 As  2 1/2x2  1/2x1 /4 

2.38 

93.750 

39.3908 

G 

1 /»2  As  3 X 3 X 6/l6 

2.38 

118.208 

49.6671 

J 

2V1  L 2 1/2  X 2 X 1/4 

3.56 

83.853 

23.5541 

1 

2^ 

1.06 

41.926 

39.5531 

2 

3 /f2  Ls  2 1/2  X 2 X 1/4 

3.56 

83.853 

23.5541 

1 

3^ 

2.12 

93.760 

44.2217 

1 

4 

2.S8 

93.750 

39.3908 

5 

4 

2.12 

83.853 

39.5531 

3 

5 All  members  having  the 

1.06 

93.750 

88.4434 

4 

5 

1.06 

93.750 

88.4434 

3 

6 same  area  have  the 

3.56 

83.853 

23.5641 

5 

6 

1.06 

41.926 

39.5531 

6 

7 same  section 

3.56 

83.853 

23.5541 

6 

7 

1.06 

93.760 

88.4434 

4 

8 

2.12 

225.000 

106.1321 

7 

9 

7 

10 

9 

10 

10 

11 

8 

9 

9 

11  Sjmimetrical  with  the 

other 

half  of  the 

Truss 

8 

11 

8 

12 

11 

12 

11 

13 

12 

13 

13 

L 

12 

K 

12 

L 

K 

L 

For  the  16-ft.,  21-ft,,  and  26-ft.  coliunn  heights  the  colnrons 
D J and  B L are  made  of  1 PI.  8 x l/4  and  4 /.s  3 x 2 l/2  x l/4 
with  the  longer  leg  outstanding.  The  moment  of  inertia  of  the 
column  section  about  a vertical  axis  normal  to  the  plane  of  the 
truss  is  81.2  inches^. 

Moment  of  inertia  for  31-ft.  column  height  is  222.0  inches^, 
*See  Fig.  26. 


T-- 


ri 


106 


TABLE  8 

DESIGN  OP  BENT 
SPAN  LENGTH  60  PT. 


Member’*'  Section 

A 

Area  in  su 

1 

.in. Length 

in.  A 

G 

J 

Same  as  column 

7,24 

72.00 

4.9448 

J 

1 

2 is  2 3/4  X 2 3/4  X 

1/4  2.62 

112.50 

42.9389 

G 

1 

2.62 

133.67 

50.9809 

J 

2 

2 is  3 1/2x3x6/16 

3.86 

100.62 

26.0674 

1 

2 

IL  2 1/4x2  1/4X1/4 

1,06 

50.31 

47.4623 

2 

3 

3.86 

100.62 

26.0674 

1 

3 

2 ts  2 1/4x2  1/4X1/4 

2,12 

112,60 

53.0660 

1 

4 

2.62 

112.60 

42.9389 

3 

4 

2.12 

100.62 

47.4623 

3 

5 

1.06 

112,50 

106.1321 

4 

5 

li.  2 1/2x2  1/2X1/4 

1.19 

112.50 

94.5378 

3 

6 

3,86 

100.62 

26.0674 

5 

6 

1,06 

50.31 

47,4623 

6 

7 

3.86 

100.62 

26.0674 

5 

7 

1.19 

112.50 

94.5378 

4 

8 

2.12 

270.00 

127.3585 

7 

9 

All  members  having 

1.19 

112.60 

94,5378 

7 

10 

3.86 

100.62 

26.0674 

9 

10 

the  same  area  have 

1.06 

50.31 

47.4623 

10 

11 

3.86 

100.62 

26.0674 

8 

9 

the  same  section 

1.19 

112.50 

94.5378 

9 

11 

1.06 

112.50 

106.1321 

8 

11 

2.12 

100.62 

47.4623 

8 

12 

2.62 

112.50 

42,9389 

11 

12 

2.12 

112.50 

53.0660 

11 

13 

3,86 

100.62 

26.0674 

12 

13 

1.06 

50.31 

47.4623 

13 

L 

3.86 

100.62 

26.0674 

12 

K 

2.62 

133.57 

50.9809 

12 

L 

2.62 

112.50 

42.9389 

K 

L 

7.24 

72.00 

9.9448 

Por  the  16-ft.,  21-ft.  and  26-ft.  coltinm  heights  the  columns 
D J and  B L are  made  of  1 PI,  8 x l/4  and  4 is  3 x 2 l/2x  l/4 
with  longer  leg  outstanding.  The  moment  of  inertia  of  the 
column  section  about  a vertical  axis  normal  to  the  plane  of  the 
truss  is  81.2  inches^.  Moment  of  inertia  for  31- ft, column 
height  is  222.0  inches^, 

* See  Pig,  26, 


TABIiE  9 

RBACTIOUS  FOR  UlTIFORM  VERTICAL  LO.U3  OF  1 LB*  PER  FOOT  LENGTH  OF  TRUSS 


106 


M 

M 

02 

02 

ra 

H 

O 


fH 


rH 


s 


p> 

h 


^ & 

02  \ 
03 

o m 


M 

CO 

02 


O 


I. 


cd  cl) 


« 

w 


<D 

02  ^ 
c6  ^ 
o w 


+5 

+> 

H r£3  ® 

;i3 

-P  -H  -P 
O ® «H 
Eh  ifl  O 


,a 

p 
to  , 

P4  ® * 

CQ  »A 


CQ  O W ^ 
CV2  00  CO 

lO  ^ ci« 

04  CO  CM  to 
CM  00  O 
^ to  'M* 

l>-  I>  04 
00  CM  rH 

^ 5^  -<vl<  ^}* 

M2  CO  O 
O M2  CO  CM 
M2  ^1* 

04  t>-  H •si* 
04  04  to  si* 
■si*  ■>;H  -M*  -st* 

CO  00  U5 
^ 00  0>  H 
00  C«-  o 

CM  O CO  t>- 
CJ4  CM  M2  CO 
O C-  CO  M2 

04  O 00  o 
CM  00  H C- 
M2  00  C- 

H H M2 
00  CO  o 
0>  H 04  H 

iH  M2  C-  si* 
00  rH  si*  04 
CM  H rH  00 

i>  »o  lO  ca 

M2  CO  CM  to 

04  M2  CO  M2 

O M2  CO  M2 
H 

02  xii  ^ 

iH 

C~  00  <£>  t- 

00  C-  H 

«x>  lO  ■sj<  -sjl 

00  M)  00  O 

to  CM  04 

M2  M2  to 

04  M2  M2  00 
M2  'M*  lO  00 
!>  M2  M2 

I>-  M2  04  M2 
l>-  ^ t> 

00  IC-  M2  M2 

00  W CM  CM 
M2  CO  CO  M2 
04  00  l>-  M2 

0>  O H lO 
CD  O O 
O to  lO  CO 
CM  H O O 

.0943 

.04267 

.02344 

.02788 

.16299 

.06842 

.03629 

.04354 

.18138 

.07516 

.03743 

.04849 

.20528 

.07947 

.04008 

.05179 

C-  00  O £> 
U2  00  *>  H 
tD  lO 

00  M2  00  O 
CO  CM  vjt  04 
M2  M2  CO 

04  M2  M2  00 
M2  M2  00 
C*-  M2  M2  -sjl 

C“  M2  04  M2 

C~  Nji  ^ ^ 
00  t>-  M2  M2 

00  to  CM  CM 
M2  CO  CO  M2 
04  00  M2 

O ■«;)<  O CM 
O CO  O 

O CM  H 
H O O O 

.0462 

.01959 

.01063 

.01382 

.07296 

.03074 

.01589 

.02088 

.07591 

.03096 

.01680 

.02180 

.08168 

.03311 

.01661 

.02348 

lO  iH  ^ 

H H CM  CM 

M2  M2  M2  *A 
• • • • 

CO  00  to  00 
CM  CM  CO  CO 

M)  H M2  H 
CM  CO  to 

M2  M2  M2  M2 

• • . . 

00  to  00  to 
CM  CO  to 

rH  M2  H M2 
to  CO  si*  -si* 

O o O O 
0 0 0*0 
U2  o 00  04 

CO  O O CO 
to  o M2  CO 
M2  00  O 

O M2  O M2 
O CM  M2  C- 
M2  M2  !>• 

O O O O 
CM  CM  CM  CM 
CO  M2  M2 

l>  O CO  c- 
M2  M2  CO  rH 
CM  CO  lO 

H 

O CM  O o> 
1 — 1 rH  rH  fH 

M2  H M2  H 
H CM  CM  to 

M2  iH  M2  H 
H CM  CM  to 

M2  H M2  H 
H CM  CM  CO 

M2  1 — 1 M2  1 — 1 
H CM  CM  to 

O O o o 

CM  CM  CM  CM 

O O o o 
CO  to  CO  CO 

o o o o 

o o o o 

M2  M2  M2  M2 

o o o o 

M2  M2  M2  M) 

107 


TABLE  10 

REACTIONS  FOR  VERTICAL  LOi-L  OP  1 LB. 
AT  POINT  1* 


Span 

Length 

L 

Column 

Height 

h 

Ratio 

h 

L 

Total  Class  I 

Height  Case  A Case  B 

of  Bent  a/t  =HR/t  V^/t 

VR/t 

30 

16 

.533 

23.5 

.001629 

.03590 

.00665 

30 

21 

.700 

28.6 

.000687 

.02961 

.00648 

30 

26 

.867 

33.6 

.000360 

.02519 

.00466 

30 

31 

1.033 

38.5 

.000478 

.02192 

.00406 

40 

16 

.400 

26 

.001875 

.03245 

.00601 

40 

21 

.525 

31 

.000773 

.02722 

.00504 

40 

26 

.650 

36 

.000415 

.02344 

.00434 

40 

31 

.775 

41 

.000541 

.02058 

.00381 

60 

16 

.320 

28.6 

.001590 

.02961 

.00548 

60 

21 

.420 

33.5 

.000655 

.02519 

.00466 

60 

26 

.520 

38.5 

.000328 

.02192 

.00406 

60 

31 

.620 

43.5 

.000453 

.01940 

.00359 

60 

16 

.267 

31 

.001417 

.02722 

.00504 

60 

21 

.350 

36 

.000573 

.02344 

.00434 

60 

26 

.433 

41 

.000291 

.02058 

.00381 

60 

31 

.517 

46 

.000410 

.01834 

.00340 

*See  Pigs  1,23, £5, and  26. 


) 


!> 


TABLE  10  (continued)  108. 


Span  Length 

L 

Column 

Height 

h 

Class 

II 

Case  A 

Case  B 

Case  C 

2L/tf=KR/t 

Vl/t 

VR/t 

%/t  %/t 

30 

16 

.003260 

.03613 

.00643 

.1704  .2508  .363 

.534 

30 

21 

.001484 

.02974 

.00535 

.1052  .1538  .347 

.508 

30 

26 

.000799 

.02628 

.00457 

.0660  .0978  .308 

.464 

30 

31 

.000969 

.02209 

.00389 

.0912  .1534  .290 

.488 

40 

16 

.004235 

.03262 

.00684 

.2041  .2844  .402 

.560 

40 

21 

.001790 

.02732 

.00494 

.1214  .1686  .377 

.623 

40 

26 

.000932 

.02351 

.00427 

.0801  .1128  .358 

.504 

40 

31 

.001128 

.02071 

.00368 

.1070  .1686  .316 

.498 

50 

16 

.003690 

.02971 

.00538 

.1949  .2549  .440 

.576 

50 

21 

.001562 

.02525 

.00460 

.1075  .1442  .382 

.513 

50 

26 

.000785 

.02196 

.00402 

.0696  .0937  .370 

.498 

50 

3lL 

.000995 

.01948 

.00351 

.1041  .1517  .349 

.508 

60 

16 

.003460 

.02729 

.00497 

.1807  .2306  .435 

.555 

60 

21 

.001371 

.02348 

.00430 

.1026  .1336  .416 

.541 

60 

26 

.000698 

.02061 

.00378 

.0638  .0840  .381 

.501 

60 

31 

.000928 

.01840 

.00334 

.0966  .1365  .347 

.490 

TABIS  11  109 

RSACTIOlfS  ITOR  VERTICAL  LOAD  OP  1 LB*  AT  POIRT  4* 


Span 

Length 

L 

Column  Ratio 
Height  h 

h L 

Total 

Class 

I 

Height 
of  Bent 
t 

Case  A 

%/t=  %/t 

Case  8. 

20^r? 

10 

.500 

15 

.008065 

.04583 

.02083 

20 

12 

.600 

17 

.004833 

.04044 

.01838 

20 

16 

.800 

21 

.002  055 

.03274 

.01488 

20 

19 

.950 

24 

.000805 

.02865 

.01302 

30 

16 

.533 

23.5 

.001886 

.02926 

.01330 

30 

21 

.700 

28.5 

.000827 

.02412 

.01096 

30 

26 

.867 

33.5 

.000443 

.02052 

.00933 

30 

31 

1.033 

38.5 

.000584 

.01786 

.00812 

40 

16 

.400 

26 

.002323 

.02644 

.01202 

40 

21 

.525 

31 

.000974 

.02218 

.01008 

40v 

26 

.650 

36 

.000600 

.01910 

.00868 

40 

31 

.775 

41 

.000651 

.01677 

.00762 

50 

16 

.320 

28.5 

.001942 

.02412 

.01096 

50 

21 

.420 

33.5 

.000782 

.02052 

.00933 

50 

26 

.520 

38.5 

.000391 

.01786 

.00812 

50 

31 

.620 

43.5 

.000555 

.01580 

.00718 

60 

16 

.267 

31 

^001691 

.02218 

.01008 

60 

21 

.350 

36 

.000687 

.01910 

.00868 

60 

26 

.433 

41 

.000351 

.01677 

.00762 

60 

31 

.517 

46 

.000500 

.01495 

.00679 

*See  Pigs.l, 

29,25,  and  26. 

4 corresponds 

to  Point 

T of  20-ft. 

span. 

See  Pig.  22. 

' '/ 


1 1 •’ 


\\" 


. V 


0 


V 


V.' 


■i 


11 


110. 

TABLE  11  (continued) 


Class  II 


Span  Column 
length  Height 
L h 

Case  A 

Case 

B 

Case  C 

®L/t=%/t 

^ L/t 

^R/t 

%/t 

%/t 

^L/d 

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20 

10 

.015720 

.04647 

.02019 

.5055 

.6591 

.447 

.582 

20 

12 

.009734 

.04091 

.01791 

.3948 

.5077 

.422 

. 54«3 

20 

16 

.004306 

.03301 

.01461 

.2473 

,3124 

.399 

.504 

20 

19 

.002644 

.02884 

.01283 

.1754 

.2224 

.369 

.467 

30 

16 

. 003906 

.02943 

,01312 

.2237 

.2866 

.398 

.510 

30 

21 

.001756 

.02423 

.01086 

.1382 

.1761 

.386 

.492 

30 

26 

.000941 

.02059 

.009257 

.0874 

.1132 

.352 

.456 

30 

31 

.001165 

.01799 

.007984 

.1226 

.1703 

.325 

.451 

40 

16 

.005131 

,02656 

.01190 

.2777 

.3350 

.451 

.544 

40 

21 

.002139 

.02225 

.01001 

.1658 

.1999 

,431 

.519 

40 

26 

.001131 

.01914 

.00863 

.1048 

.1275 

.386 

.470 

40 

31 

.001374 

.01686 

.00753 

.1513 

.1946 

.367 

.472 

50 

16 

.004696 

.02420 

.01088 

.2496 

.2981 

.443 

,52y 

60 

21 

.001880 

.02057 

.00928 

.1393 

.1697 

.411 

.501 

50 

26 

.000957 

.01789 

.00808 

.0884 

.1087 

.385 

.473 

50 

31 

.001238 

.01587 

.00712 

.1347 

.1733 

.363 

.467 

60 

16 

.004346 

.02223 

.01003 

,2398 

.2793 

.460 

. 53t> 

60 

21 

.001665 

.01913 

.00865 

.1329 

.1569 

.444 

.624 

60 

26 

.000852 

.01679 

.00760 

.0805 

.0963 

,394 

.471 

60 

31 

.001096 

.01499 

.00675 

.1256 

.1569 

.382 

.477 

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